text:
# General parameterisation of the geodesic equation

1. Mar 14, 2014

### victorvmotti

Hello all,

In Carroll's on page 109 it is pointed out that for derivation of the geodesic equation, 3.44, a "hidden" assumption is that we have used an affine parameter.

Some few lines below we see that "any other parametrization" could be used, called alpha, but in that case the general form of geodesic equation will be 3.58 and not 3.44.

However, when plugging 3.59 into 3.58 and rearranging I can only recover the geodesic equation, 3.44, provided that alpha is itself a linear function of λ, and not "some general parameter alpha."

So unlike what is said below 3.59, we cannot "always" find an affine parameter λ.

Also, related to this confusion is what we call "a null geodesic."

To make it clear for myself I need to see an example of a null path which is NOT a geodesic.

Can someone please introduce or describe such a null path?

2. Mar 14, 2014

### Bill_K

How about a null spiral? A particle travels endlessly in a circular path, with tangential velocity c.

x = R cos Ωt
y = R sin Ωt
z = 0

with RΩ = c

3. Mar 14, 2014

### stevendaryl

Staff Emeritus
I don't have Carroll's book, but unless I made a mistake, here's the general equation:

$g_{\mu \nu} A^\nu + \Gamma_{\mu \nu \lambda} U^\lambda U^\nu = g_{\mu \nu} U^\nu \dfrac{d}{ds} (ln \mathcal{L})$

where $s =$ the parameter,
$g_{\mu \nu} =$ the metric tensor,
$U^\mu = \dfrac{d}{ds} x^\mu(s)$
$A^\mu = \dfrac{d}{ds} U^\mu(s)$
$\mathcal{L} = \sqrt{g_{\mu \nu} U^\mu U^\nu}$
$\Gamma_{\mu \nu \lambda} =$ the connection coefficients, computed from derivatives of $g_{\mu \nu}$

If you choose the parameter $s$ so that $\mathcal{L}$ is constant, then the right-hand side is zero, and you get the usual form of the geodesic equation.

To make $\mathcal{L}$ constant, you just choose $s$ to satisfy:

$ds = \sqrt{g_{\mu \nu} dx^\mu dx^\nu}$

which is proper time (or any other linearly related parameter).

This only works for non-null geodesics. If we're talking about a null geodesic, then $\mathcal{L}$ is identically zero, so I'm not sure what the condition is on the parametrization.

4. Mar 14, 2014

### pervect

Staff Emeritus
I'm not quite seeing an equation # match to the online version of Caroll at http://arxiv.org/pdf/gr-qc/9712019v1.pdf or http://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html

This online version seems much clearer than I remember it being, perhaps Caroll has revised it?

Anyway, a quote to enable one to find the section in question, which might be helpful.

and

5. Mar 14, 2014

### stevendaryl

Staff Emeritus
My equation for a general parametrization is basically the same as Carroll's equation 3.59, except that the arbitrary function is explicitly written as $\dfrac{d}{ds} (ln \mathcal{L}$. In that form, we can easily see that the right-hand side vanishes if you choose a linear function of proper time. The function $\mathcal{L}$ is just $\dfrac{d \tau}{ds}$ where $\tau$ is proper time. So if $\tau$ is a linear function of $s$, then that's a constant, and so is $ln (\dfrac{d \tau}{ds})$, and so its derivative is zero.

6. Mar 15, 2014

### victorvmotti

Hello Again,

Thanks for the feedback.

What is still unclear for me is where we have made the implicit assumption of an affine parameter to derive the usual form of the geodesic equation.

We start with parallel transport of the tangent vector to the path and demand that its directional covariant derivative vanishes. This gives the usual form of the geodesic equation.

So the question is that where we have derived from the general parametrization of the geodesic equation in the first place?

Also, Carroll says that for any parameter in the general form you can always find an affine parameter to satisfy the usual form of the geodesic equation. But how?

About null paths that are not geodesic is it possible to illustrate that an endlessly circular path does not satisfy the geodesic equation?

But then what is the geodesic equation for a null path?

Should we use the form based on the four-momentum that is derived and used for timelike paths?

7. Mar 15, 2014

### pervect

Staff Emeritus
I thought the section I quoted from the online version of caroll was pretty clear. A geodesic can be regarded as a curve that parallel transports itself (and this appears to be the preferred definition for mathemeticians). If you only require that a curve parallel transport itself in the same direction, allowing parallel transport to change the length of a vector, there is no requirement on how you parameterize the curve doing the transporting. When you require that parallel transport not change the length of a transported vector, there is an implied requirement as to how the transporting curve must be parameterized.

This requirement of unchanging length of the tangent vector can be used for timelike, spacelike, or null geodesics. Parameterizing by proper time is a special case that applies only to timelike geodesics.

To go a bit beyond Caroll and provide a physical intuitive reaso for the affine parameteriztion of null geodesics (rather than a formal mathematical one)

If you consider an actual plane electromagnetic wave following a null geodesic, the EM wave will have regularly spaced nulls where the electric and magnetic fields both vanish. I.e. for a plane EM wave the electric field is E sin (omega t) and the magnetic field B sin( omega t). At certain times E and B are both zero, creating a null. These nulls must be equally spaced if the geodesic is parameterized affinely.

8. Mar 15, 2014

### WannabeNewton

The geodesic equation under any parametrization reads $\xi^{\nu}\nabla_{\nu}\xi^{\mu} = \alpha \xi^{\mu}$ for some smooth scalar field $\alpha$ defined on the geodesic. So what the geodesic equation under any parametrization really says is the tangent vector's direction is parallel transported along the geodesic but the length isn't. This is easy to see as $\xi^{\nu}\nabla_{\nu}\frac{\xi^{\mu}}{(-\xi_{\gamma}\xi^{\gamma})^{1/2}} = \frac{\alpha \xi^{\mu}}{(-\xi_{\gamma}\xi^{\gamma})^{1/2}} + \frac{\xi_{\gamma}\xi^{\nu}\nabla_{\nu}\xi^{\gamma}}{(-\xi_{\gamma}\xi^{\gamma})^{3/2}} = 0$.

However it is also easy to show that given the above, one can always reparametrize the geodesic so as to satisfy $\xi^{\nu}\nabla_{\nu}\xi^{\mu} = 0$. This is called an affine parametrization.

See here: https://www.physicsforums.com/showpost.php?p=4381515&postcount=4

9. Mar 15, 2014

### stevendaryl

Staff Emeritus
There are two ways to go about deriving the geodesic equation. They are equivalent for the connection used in General Relativity.

One approach is to maximize proper time:

$\tau = \int \mathcal{L} ds$

where $\mathcal{L} = \dfrac{d\tau}{ds} = \sqrt{g_{\mu \nu} U^\mu U^\nu}$
and where $U^\mu = \dfrac{dx^\mu}{ds}$

Using the calculus of variations, we maximize proper time for a path $x^\mu(s)$ satifying:

$\dfrac{d}{ds} (\dfrac{\partial \mathcal{L}}{\partial U^\mu})= \dfrac{\partial}{\partial x^\mu} \mathcal{L}$

This doesn't assume anything about the parameter $s$ being affine, but the resulting equation is only the usual geodesic equation in the case where the parameter is affine.

The other approach to deriving the geodesic equation is to start with the vector equation

$\dfrac{D}{D s} U = \alpha U$

where $U$ is the velocity 4-vector, and where $\alpha$ is an arbitrary function of $s$ and where $\dfrac{D}{D\tau}$ is the operator defined by the following procedure: (I can never remember what the name of it is--it's not a covariant derivative, but it's related to it)

$(\dfrac{D}{D s} A)^\mu = \dfrac{d A^\mu}{ds} + \Gamma^\mu_{\nu \lambda} A^\nu U^\lambda$

The two approaches give the same equation, except in the first approach, the function $\alpha$ is not arbitrary, but is equal to $\dfrac{d}{ds}(ln \mathcal{L})$.

These are actually two different things: an extremal path, versus a path whose velocity vector is unrotated by parallel transport, but they turn out to be equal in GR.

10. Mar 15, 2014

### stevendaryl

Staff Emeritus

$\dfrac{d^2}{ds^2} x^\mu + \Gamma^\mu_{\nu \lambda} \dfrac{dx^\nu}{ds} \dfrac{dx^\lambda}{ds} = \alpha \dfrac{dx^\mu}{ds}$

Now, change parameters to $\tau$. Let $Q = \dfrac{d \tau}{d s}$
The equation becomes (after canceling factors of $Q$ from both sides):

$\dfrac{1}{Q} \dfrac{d Q}{d\tau} \dfrac{dx^\mu}{d \tau} + \dfrac{d^2}{d\tau^2} x^\mu + \Gamma^\mu_{\nu \lambda} \dfrac{dx^\nu}{d\tau} \dfrac{dx^\lambda}{d\tau} = \dfrac{\alpha}{Q} \dfrac{dx^\mu}{d\tau}$

So if you choose $Q$ to satisfy

$\dfrac{d Q}{d \tau} = \alpha$

or, in terms of $s$

$\dfrac{1}{Q} \dfrac{d Q}{d s} = \dfrac{d (ln Q)}{ds} = \alpha$

then you get the usual form of the geodesic equation.

Sure, you just plug it into the geodesic equation, and show that it doesn't work.

The second approach to deriving a geodesic equation works exactly the same whether or not the path is null. So the same equation applies. The first approach (extremizing proper time) doesn't work for null geodesics, since every null path has the same proper time, zero.

11. Mar 15, 2014

### victorvmotti

Seems that Carroll calls it directional covariant derivative.

12. Mar 15, 2014

### victorvmotti

Doesn't this imply that alpha should be zero given how we defined Q above?

You seem to be so close to what Carroll has written in the book, that is relating the general parameter to an affine parameter, using your equation above I can recover the equation he suggests to derive a usual form of the geodesic from the general form, but a difference here is that he says alpha, used in your general form, should be a function of s.

Last edited: Mar 15, 2014
13. Mar 15, 2014

### victorvmotti

This was particularly helpful to make sense of a parallel transport. Because Carroll, when introducing the parallel transport, demands the norm of the vector, or in general "the components" of an arbitrary rank tensor, should not change which means that its "directional covariant derivative" should vanish.

When you say that fundamental to parallel transport notion is that "velocity vector is unrotated" the general parametrization of the geodesic equation makes a lot more clear sense.

14. Mar 16, 2014

### victorvmotti

If we are in locally inertial coordinates then connection coefficients vanishes and therefore both null and timelike geodesic paths should be described by coordinates that are exponential functions of the general parameter s. Am I correct?

If we are in general coordinates then on null paths the term involving connection coefficients in the general form of the geodesic equation vanishes because the inner product of vectors on a null path is zero. So again the null geodesic path is described by coordinates that are exponential functions of the general parameter s. Am I correct?

15. Mar 16, 2014

### stevendaryl

Staff Emeritus
Oh, I didn't actually describe the procedure, I just gave the equation. The procedure is this, in a little more detail:

1. Start with a parametrized path $\mathcal{P}(s)$.
2. Let $A(s)$ be an arbitrary 4-vector that is defined along the path $\mathcal{P}(s)$
3. Let $\tilde{A}(s + \delta s)$ be the result of parallel-transporting $V(s + \delta s)$ back along the path $\mathcal{P}$ from the point $\mathcal{P}(s + \delta s)$ to the point $\mathcal{P}(s)$
4. Let $\delta A = \tilde{A}(s+\delta s) - A$
5. Then $\dfrac{D}{Ds} A = {lim}(\delta s \rightarrow 0) \dfrac{\delta A}{\delta s}$

16. Mar 16, 2014

### stevendaryl

Staff Emeritus
The connection terms don't vanish for a null path. Let's look at a specific simple example: cylindrical coordinates: $(t, z, \rho, \phi)$

The metric components are:
$g_{tt} = 1$
$g_{zz} = g_{\rho \rho} = -1$
$g_{\phi \phi} = - \rho^2$

The non-zero connection coefficients are (if I computed them correctly)
$\Gamma^\rho_{\phi \phi} = \rho$
$\Gamma^\phi_{\rho \phi} = \Gamma^\phi_{\phi \rho} = \dfrac{1}{\rho}$

So the geodesic equation is (in components):

$\dfrac{d^2 \phi}{ds^2} + \dfrac{2}{\rho}\dfrac{d \phi}{ds} \dfrac{d \rho}{ds} = \alpha \dfrac{d \phi}{ds}$

$\dfrac{d^2 \rho}{ds^2} + \rho (\dfrac{d \phi}{ds})^2 = \alpha \dfrac{d \rho}{ds}$

$\dfrac{d^2 z}{ds^2} = \alpha \dfrac{d z}{ds}$

$\dfrac{d^2 t}{ds^2} = \alpha \dfrac{d t}{ds}$

Being a null geodesic means that
$(\dfrac{dt}{ds})^2 - (\dfrac{dz}{ds})^2 - (\dfrac{d\rho}{ds})^2 - \rho^2 (\dfrac{d\phi}{ds})^2 = 0$

17. Mar 16, 2014

### victorvmotti

Actually I meant this, not sure if I am correct:

$g_{\nu \lambda}\dfrac{d^2}{ds^2} x^\mu + \Gamma^\mu_{\nu \lambda}g_{\nu \lambda} \dfrac{dx^\nu}{ds} \dfrac{dx^\lambda}{ds} = \alpha g_{\nu \lambda}\dfrac{dx^\mu}{ds}$

Null path, then

$g_{\nu \lambda} \dfrac{dx^\nu}{ds} \dfrac{dx^\lambda}{ds} = 0$

$g_{\nu \lambda}\dfrac{d^2}{ds^2} x^\mu = \alpha g_{\nu \lambda}\dfrac{dx^\mu}{ds}$

$\dfrac{d^2}{ds^2} x^\mu = \alpha\dfrac{dx^\mu}{ds}$

$x^\mu = \alpha\exp{(s\alpha)}$

18. Mar 16, 2014

### stevendaryl

Staff Emeritus
That is not the correct equation. The geodesic equation is:

$\dfrac{d^2}{ds^2} x^\mu + \Gamma^\mu_{\nu \lambda} \dfrac{dx^\nu}{ds} \dfrac{dx^\lambda}{ds} = \alpha \dfrac{d x^\mu}{ds}$

Typically in equations involving indices, there should never be repetitions of indices in a term except in the case of one raised index and the corresponding lowered index. In your equation, the indices $\nu$ and $\lambda$ on $\Gamma^\mu_{\nu \lambda} \dfrac{dx^\nu}{ds} \dfrac{dx^\lambda}{ds}$ are dummy indices. They should be replaced by $\nu'$ and $\lambda'$ to avoid being confused with the indices on $g_{\nu \lambda}$. If you do that, you'll see that $g_{\nu \lambda}$ has no effect, and you can just divide through by it.

Last edited: Mar 16, 2014
19. Mar 16, 2014

### victorvmotti

Thanks, you know I multiplied both sides of the general form the geodesic equation by the metric to cancel the connection term. I learned here that it is not allowed if we do not change the indices of the metric tensor.

But still the general form of the geodesic equation, I mean when we are not using an affine parameter, makes me confused.

Suppose when we are in Riemann normal coordinate, or locally inertial coordinates,

For both timelike and null geodesic paths we have

$\dfrac{d^2}{ds^2} x^\mu + \Gamma^\mu_{\nu \lambda} \dfrac{dx^\nu}{ds} \dfrac{dx^\lambda}{ds} = \alpha \dfrac{dx^\mu}{ds}$

Assuming a Riemann normal coordinate then $\Gamma^\mu_{\nu \lambda}$ vanishes and then

$\dfrac{d^2}{ds^2} x^\mu = \alpha \dfrac{dx^\mu}{ds}$

$x^\mu = \alpha\exp{(s\alpha)}$

Right?

20. Mar 16, 2014

### stevendaryl

Staff Emeritus
Sort of. I think you mean:

$x^\mu = a^\mu \exp{(s\alpha)} + b^\mu$

where $a^\mu, b^\mu$ are constant vectors. That's the general solution to the equation

$\dfrac{d^2}{ds^2} x^\mu = \alpha \dfrac{dx^\mu}{ds}$

But locally inertial coordinates are only good locally. So that solution is only valid for small values of $s$.


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# Why 'Mode()[0]', Not Just 'Mode()?

In 7/9 stats Module of "mode. The output of mode : 0 6. Here, is that true? 0 refers to index and 6 is the mode value?

my original code " houses[‘Mo Sold’].mode() " did not work.
The error says ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
Only I changed the code as the solution’s— mode()[0], then it works.

I did notice in our FAQ discussion forum, someone asked the same question and whoever gave the answer didn’t clarify my confusion. Do we always have to have mode()[0]? I checked geeksforgeeks or other documentation, it didn’t suggest mode()[0].

import pandas as pd
help(pd.Series.mode)

mode(self, dropna=True)
Return the mode(s) of the dataset.

Always returns Series even if only one value is returned.

Parameters
----------
dropna : bool, default True
Don't consider counts of NaN/NaT.

Returns
-------
Series
Modes of the Series in sorted order.

x = houses["Mo Sold"].mode()

.mode() returns a series regardless whether one value is returned.

The series for houses["Mo Sold"].mode():

\begin{array}{|c|c|} \hline - & 0 \\ \hline 0 & 6\\ \hline \end{array}

houses["Mo Sold"].mode()[0] returns the single value in the series.

1 Like

There can be multiple modes.

x = [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3]

Then there will be three modes with frequency 4 - 1, 2, and 3.

>>> import pandas as pd

Sample Series data:

>>> s = pd.Series([1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3])

>>> s
0     1
1     1
2     1
3     1
4     2
5     2
6     2
7     2
8     3
9     3
10    3
11    3
dtype: int64

Using .mode:

>>> x = s.mode()
>>> x
0    1
1    2
2    3
dtype: int64

Retrieving multiple modes:

>>> x[0]
1
>>> x[1]
2
>>> x[2]
3

1 Like

@alvinctk Thank you for your explanation. It does explain this question very well. But here comes a new question.
So the output(in my picture) of "print(mode) is 0 and 6 means what? two modes or 0 refers to index and 6 is mode?

I am confused of this picture you showed me. Is it a series or an array? 0 on the top right means column index and 0 on the bottom left means row indexing?

I did notice in stats course, I have to be careful of differentiating list/array and series because they represent different way to write the function. eg .series.sum() or sum(array).

0 is just to indicate the indexes. It is a series.
You can see it on dataquest variable checker.
x = houses["Mo Sold"].mode()

1 Like

Pandas follows the numpy convention of raising an error when you try to convert something to a bool. This happens in a if or when using the boolean operations, and, or, or not. It is not clear what the result of.

example

5 == pd.Series([12,2,5,10])

The result you get is a Series of booleans, equal in size to the pd.Series in the right hand side of the expression. So, you get an error. The problem here is that you are comparing a pandas pd.Series with a value, so you’ll have multiple True and multiple False values, as in the case above. This of course is ambiguous, since the condition is neither True or False. You need to further aggregate the result so that a single boolean value results from the operation. For that you’ll have to use either any or all depending on whether you want at least one (any) or all values to satisfy the condition.

(5 == pd.Series([12,2,5,10])).all()
# False

or

(5 == pd.Series([12,2,5,10])).any()
# True

1 Like

format: plain


text:
5

# 37. What chromosome aberration do individuals with familial Downsyndrome exhibit?Multiple ChoiceChromosome 21 duplicationChromosome 14 trisomyChromosome fusion invo...

## Question

###### 37. What chromosome aberration do individuals with familial Downsyndrome exhibit?Multiple ChoiceChromosome 21 duplicationChromosome 14 trisomyChromosome fusion involving 14 and 21Reciprocal balanced translocation between chromosomes 14 and 21

37. What chromosome aberration do individuals with familial Down syndrome exhibit? Multiple Choice Chromosome 21 duplication Chromosome 14 trisomy Chromosome fusion involving 14 and 21 Reciprocal balanced translocation between chromosomes 14 and 21

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text:
# Convergent or Divergent?

1. May 7, 2012

### joe_cool2

I am to find whether the sum of (n!)/(n^n) converges or diverges. I tried both the limit comparison test, and a regular comparison test. (These are the only types of tests I am allowed to use.) So I tried several approaches:

Approach #1: (n!)/(n^n) > 1/(n^n)

Normally we use a setup like this to prove something with a p-series. However, the expression on the left side of the inequality isn't a p-series.

Approach #2: (n!)/(n^n) < n!

While this expression is true, it is not useful because the formula for the series is less than, not greater than, the series that is known to diverge.

Approach #3 (Limit comparison): an = (n!)/(n^n) ; bn = 1/(n^n)

an/bn = n!

The limit here is, unfortunately, infinite, and I have to stop here.

What other approach can I take that would result in more success?

2. May 7, 2012

### chiro

Hey joe_cool2 and welcome to the forums.

My intuitive guess is that your function will converge. One test to show this that seems appropriate would be the ratio test. Take a look at this page:

http://en.wikipedia.org/wiki/Ratio_test

3. May 7, 2012

### joe_cool2

Hello, thanks for the welcome. It is much appreciated.

I am perfectly aware that it is often wise to use the ratio test in these situations with n!. However, I have been restrained to using specifically those two techniques mentioned earlier. Any ideas?

4. May 7, 2012

### micromass

Staff Emeritus
The idea is to compare your function to a geometric series. That is, you have to find an $0\leq a<1$ such that

$$\frac{n!}{n^n}\leq a^n$$

for all n. If you can find such an a, then your series will converge. So we need to find an a such that

$$\frac{n!}{(an)^n}<1$$

Try to show that the left hand side is decreasing from a certain point on.

5. May 7, 2012

### sharks

Whenever factorial is involved in series, the ratio test is your best bet.

6. May 7, 2012

### micromass

Staff Emeritus
Can you show that $\sum\frac{n^n}{(n!)^n}$ converges? If you can, then use the limit comparison test with that and your series.


text:
Challenge

You must output the current time continuously (until cancelled by an interrupt), once every second, by any of the following means:

• It must be in 24-hour or AM/PM format.
• If it is the former, it must be spaced out with colons (i.e. 15:47:36).
• If it is the latter, it must be spaced out with colons and have the AM/PM following (i.e. 3:47:36 PM)
• It may be pulled from the internet.
• It may be the system time.
• It must output any naturally accessible form of output which supports text that you choose.
• Output may have extra information aside of the time in it, but you must guarantee one, and only one, output of time per second.
• The continuous output must be a second apart - if you choose to wait until the second changes between outputs, that is fine. If you wait a second between each output, that is perfectly acceptable, despite the eventual loss of accuracy.

Since this is a catalog, languages created after this challenge are allowed to compete. Note that there must be an interpreter so the submission can be tested. It is allowed (and even encouraged) to write this interpreter yourself for a previously unimplemented language. Other than that, all the standard rules of must be obeyed. Submissions in most languages will be scored in bytes in an appropriate preexisting encoding (usually UTF-8).

Catalog

The Stack Snippet at the bottom of this post generates the catalog from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.

## Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

## Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

## Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the snippet:

## [><>](http://esolangs.org/wiki/Fish), 121 bytes

var QUESTION_ID=65020,OVERRIDE_USER=44713;function answersUrl(e){return"//api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"//api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="
"+e.body.replace(OVERRIDE_REG,"")+"
")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
body{text-align:left!important}#answer-list,#language-list{padding:10px;float:left}table thead{font-weight:700}table td{padding:5px}
 <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b">

Leaderboard

	Author	Language	Size

Winners by Language

Language	User	Score

{{PLACE}}	{{NAME}}	{{LANGUAGE}}	{{SIZE}}	Link

{{LANGUAGE}}	{{NAME}}	{{SIZE}}	Link

• Does "until cancelled by an interrupt" include closing the program or does it need a mechanism to stop the cycle coded into it? – user81655 Nov 27 '15 at 23:00
• – Mego Nov 28 '15 at 6:21
• Warning: All sleep 1 based answer break rule 5: you must guarantee one, and only one, output of time per second. !! – F. Hauri May 17 '16 at 23:03
• F. Hauri: How so? Rule 6 says "If you wait a second between each output, that is fine as well." – YetiCGN Aug 26 '16 at 22:03
• @Jasen No, but the meaning of "that is fine" implies that the wait option is still valid, regardless of the eventual time loss. I have clarified this for you. – Addison Crump Jan 7 '17 at 19:07

Japt, 8 bytes

Japt is a shortened version of JavaScript.

1e3i@OpÐ

Test it online!

How it works

1e3i@       // Repeat this function every 1e3 milliseconds:
OpÐ    //  Output new Date(), followed by a newline.

Note that this also outputs a seemingly random integer when the code is run. To get rid of this, use this code:

1e3i@OpÐ};P

On my computer, the time is formatted as 2015-11-28T16:00:18.013Z. If you don't like that, try this code instead:

1e3i@OpÐ s8};P

Which will lovingly print your time in the format 11:00:18 AM.

K, 25 24 Bytes

Removed semicolon from the end of the func!

      .z.ts:{-1@$18h$x}
\t 1000
13:11:44
13:11:45
13:11:46
13:11:47
13:11:48
...

.z.ts is called every 1000 milliseconds.
x is the time.
18$x casts the time to the appropriate format.$ - strings the result.
-1@(stringed result) - prints the string to the console.

*Edit Below is less but I wasn't sure whether the time format was allowed;

    .z.ts:{-1($x)} \t 1000 2016.08.01D00:35:37.392683000  C/C++, 219 Bytes Here is a first pass attempt. It is written in C++ but uses only C functionalities, so it could be used as C code with only changes to the include statements (I think). #include<stdio.h> #include<ctime> time_t r;struct tm*x;int main(){char c[9],b[9];time(&r);x=localtime(&r);strftime(c,9,"%T",x);puts(c);for(;;){time(&r);x=localtime(&r);strftime(b,9,"%T",x);if(b[7]!=c[7]){break;} }main();}  I think everything is pretty straightforward. It takes the time at the beginning of the main function, then it continues getting time until one second has passed. It then calls itself. Ungolfed: #include<stdio.h> #include<ctime> time_t r;struct tm*x; int main(){ char c[9];char b[9]; time(&r); x=localtime(&r); strftime(c,9,"%T",x); puts(c); for(;;){ time(&r); x=localtime(&r); strftime(b,9,"%T",x); if(b[7]!=c[7]){break;} // break when the 'second' changes } main(); }  • Erm... I'm not the best with C, but won't this cause a StackOverflow (if I remember right, this might be java seeping in)? – Addison Crump Nov 28 '15 at 1:41 • It doesn't on my computer (using g++) although int main(){main();} does cause a seg fault. – Liam Nov 28 '15 at 1:42 • @VoteToClose I believe that tail recursion gets optimized away. – Maltysen Nov 28 '15 at 1:46 • I get a segfault at about 40000 lines in (removed the waiting system). Don't think it particularly matters at that far in, though. – Addison Crump Nov 28 '15 at 1:50 • With the -O2 switch, gcc/g++ will perform tail call optimization. I'm not sure that's strictly necessary for this answer (since VTC already said it's fine as-is), but it's helpful to know. :) – Mego Nov 28 '15 at 2:09 Vimscript, 44 bytes while 1 echo strftime("%T") sleep 1 endwhile  Run like so: vim -c ":so FILE"  Pyth, 19 16 bytes It takes the relevant part of the datetime list and joins by colons. Then it passes till the second changes, with the entire thang wrapped in an infinite loop. #j\:KP.d2WqeK.d8  While it obviously doesn't work online, you can check its main idea here. • @VoteToClose oh, I actually wrote that a couple mins before you posted. Figured it was okay since it was a catalog. – Maltysen Nov 27 '15 at 23:00 • @FryAmTheEggman That's not necessary, as proven by the Stuck answer and the TI-BASIC answer. It is still human-readable and it is the time. – Addison Crump Dec 1 '15 at 11:00 • Also, you don't need to cut it for just the time string. ;P – Addison Crump Dec 1 '15 at 11:00 PHP, 33 bytes <?for(;;)sleep(print date("r "));  Output is similar to Sat, 28 Nov 2015 11:25:16 +0700, i.e. RFC 2822. • Not sure wether this is allowed. This doesn't output every second for me. This only creates a time value once per second, but doesnt actually show me. – Martijn Nov 30 '15 at 14:24 • Are you invoking from the command line, e.g. php time.php? – primo Nov 30 '15 at 16:36 Mathematica, 41 36 bytes While[1>0,Echo@DateString[];Pause@1]  Works as a script. CJam, 21 bytes L{et6<':*_@={_oNo}|}h  Try it online! Note that the online interpreter will kill the program after one minute. Same idea, but with pretty output: L{et6<3>"%02d:"fe%sW<_@={_oNo}|}h  Try it online! Go, 6469 79 bytes Go isn't very golfy. package main;import(."fmt";."time");func main(){for{Println(Now());Sleep(1e9)}}  You could try it here or on another online interpreter however, for{} is an infinite loop and those are disallowed for obvious reasons. I haven't actually tested thisthere are no go interpreters for my android ): but it should work in theory. Running it locally might cause your CPU fan to speed up, but it won't hog memory. If it crashes with some sort of OOM-error or panic: runtime out of bound, just change it to for x:=0;x<9999;x++{ ... }. • This sleeps for one nanosecond, which results in multiple outputs per second. You'll need Sleep(Second) to fix this. – Michael Hampton Nov 28 '15 at 8:55 • Sleep(1e9) would work, no? I don't remember very much about Go, but I remember something like that. – Addison Crump Nov 28 '15 at 11:05 • @VoteToClose Yes, 1e9 works fine on my cgo 1.5.1... – Michael Hampton Nov 28 '15 at 14:05 • Woo, helped golf a language I've never used! \o/ – Addison Crump Nov 28 '15 at 14:06 • Well yes, that too. I'm using syntastic and vim-go, which has about the same result... – Michael Hampton Nov 28 '15 at 14:12 SpecBAS - 45 bytes Prints in the upper left corner for a continuously updating 24 hour clock. Pressing Esc stops it running. 1 TEXT AT 0,0;TIME$(TIME,"hh:mm:ss"): GO TO 1

• I don't know if this answer would be shorter, but you don't have to necessarily just have the time string. If you output a time string with extraneous data (but still containing the time string), that is fine, so you might have a shorter answer in store. – Addison Crump Nov 28 '15 at 14:46
• TIME on it's own returns a decimal number (days since 30/12/1899!) which then needed formatting to hours/min/secs format. There isn't a function/statement that just returns already formatted date/time string unfortunately. – Brian Nov 28 '15 at 15:04

Processing, 68 bytes

void draw(){frameRate(1);println(hour()+":"+minute()+":"+second());}

Easy one I have to admit. In processing there is the draw() method which is called at a specific frequency that can be modified during the execution.

• Can you show an example output? :D – Addison Crump Nov 28 '15 at 21:40
• A new line every seconds (represented by the spaces) 22:42:32 22:42:33 22:42:34 22:42:35 22:42:36 22:42:37 – 6infinity8 Nov 28 '15 at 21:44
• Oh, so it's not graphical output? – Addison Crump Nov 28 '15 at 21:45
• No, that's printed in the console. I'm just "exploiting" the draw method as a loop. – 6infinity8 Nov 28 '15 at 22:13
• Ohh, I see. Well done, +1 (in about an hour - can't vote right now. :c) – Addison Crump Nov 28 '15 at 22:13

Ruby, 25 24 bytes

Thanks @manatwork for golfing off a byte :P

loop{putsdate;sleep 1}

• print? Either puts or $><< are shorter with identical result. – manatwork Nov 29 '15 at 15:29 • Derp. Thanks for that. (You can tell I don't use Ruby much :P) – a spaghetto Nov 29 '15 at 15:55 Ceylon, 126 112 bytes import ceylon.time{now}void p(){for(e in{now().time().string[0:8]}.cycled.paired){if(e[0]!=e[1]){print(e[1]);}}} This uses the ceylon.time library for getting the current time (in the default timezone) and formatting it (the .string function for time outputs something like 22:33:45.234, so I just take the first 8 characters of it). Unfortunately, there seems to be no sleep function in Ceylon (because it is not easy to implement in JavaScript), therefore I'm doing a busy loop here, comparing each formatted string to the previous one and printing only when there is a change. Here is a formatted version of the second version, which uses a for-loop over an infinite iterable: // print the current time (each second). // // Question: http://codegolf.stackexchange.com/q/65020/2338 // My answer: http://codegolf.stackexchange.com/a/65130/2338 import ceylon.time { now } void p() { for(e in { now().time().string[0:8] }.cycled.paired) { if(e[0]!=e[1]) { print(e[1]); } } }  Using a < instead of != would save a character, but then the program would stop working as soon as the clock hits midnight. This was the original, more functional, version resulting in 112 characters: import ceylon.time { now } void p() { { now().time().string[0:8] } .cycled .paired .map((e) => e[0] != e[1] then e[1]) .coalesced .each(print); }  • Is n=now saving you bytes if you only use it once? I don't know Ceylon, but it seems redundant. – cat Nov 29 '15 at 15:58 • @sysreq actually, no ... for this three-letter identifier it doesn't safe anything (though it also doesn't cost bytes). I'll change this. – Paŭlo Ebermann Nov 29 '15 at 19:02 Hassium, 43 Bytes func main(){print(time())sleep(1000)main()}  See expanded here sh (+ top), 7 bytes Script (Debian8): top -d1  Script (NetBSD7): top -s1  (both without trailing newline) Output (1st line only, Debian8): top - 16:25:15 up 3 days, 8:56, 4 users, load average: 0,97, 1,10, 1,03  Output (1st line only, NetBSD7, different time zone): load averages: 1.46, 1.28, 1.22; (SPACES) up 44+08:52:28 (SPACES) 15:22:06  The amount of spaces depends on the terminal width. C, 74 73 bytes #include<time.h> main(){for(time_t t;!sleep(1);time(&t))puts(ctime(&t));}  The code can work even without #include <stdio.h> and #include <unistd.h>, see live code example on ideone.com -- note: the sleep is zero there, to generate some output before the time limit has been exceeded. The first output line should be ignored ("Extra output doesn't really matter, as long as the program at some point continuously outputs time."). PureBasic, 64 59 bytes Not much to say, infinite loop, output time to the debug window with a 1000 millisecond delay a: Debug FormatDate("%hh:%ii:%ss",Date()) Delay(1e3) Goto a  • ForEver? Seems legit. – Addison Crump Dec 1 '15 at 10:24 • the comment on ForEver made me think about infinite loops and all it boils down to is a Goto loop and it saved me 5 bytes :D – Fozzedout Dec 2 '15 at 11:57 • Goto? Even better. ;D Can you put the a: directly before Debug... to remove the newline? Or is that a different version of BASIC? – Addison Crump Dec 2 '15 at 11:58 • nope, labels for goto's have to be the only statement on the line for it to work. The colon is generally used for stringing multiple commands on one line, so you could have "Debug FormatDate("%hh:%ii:%ss",Date()):Delay(1e3):Goto a" – Fozzedout Dec 3 '15 at 14:42 Prolog (SWI), 72 bytes Formating to user_output stops working when using sleep, so I had to save the time to a variable and print it explicitly. Code: p:-repeat,get_time(T),format_time(atom(X),'%T',T),write(X),sleep(1),1=0.  Explained: p:-repeat, % loop until success get_time(T), % Get current timestamp format_time(atom(X),'%T',T), % Format timestamp as HH:mm:ss write(X), % Print time sleep(1), % Sleep 1 second 1=0. % Fail and go back to repeat  Vitsy + bash, 13 11 bytes This language feature was made after this question, but not for this question. <w1Z,'date' < Loop leftwards in the code. Rest of code is written in reverse for "readability". 'etad' Push 'date' to the stack. , Do the shell script for what's in the stack. Z Output the result. 1w Wait a second.  Or, using eval... <w1Zn"Date()" < Go leftwards through this code in order to loop infinitely. Rest of code is written in reverse for "readability". ")(etaD" Push "Date()" to the stack. n Eval through JS. Z Output everything. 1w Delay for one second.  Try it online! • sleep 1 based answer break rule 5: you must guarantee one, and only one, output of time per second... anymore!! – F. Hauri May 18 '16 at 6:44 • @F.Hauri Please read rule 6 - also, stop spamming me. – Addison Crump May 18 '16 at 6:45 Mouse-2002, 84 bytes Mouse is cool, but it's not the golfiest stack based language on the planet (well, not until I update it in a reimplementation -- then it will be better.) 0&FIX 2&WIDTH "!"(&HOUR &!DEC ":"&MIN &!DEC ":"&SEC &!DEC 8!'8!'8!'8!'8!'8!'8!'8!')$

The 8!' prints the character with ASCII code 8, or backspace. You probably need an ANSI terminal to run this; it updates in place by backspacing over itself and writing the new time once per second.

sample use:

$mouse clock2.mou 19:47:34  Based on this, but golfed. Python 2, 85 bytes import time;while 1:print __import__('datetime').datetime.now().time();time.sleep(1)  import time;time.sleep() has 2 fewer bytes than __import__('time').sleep(), while import datetime;datetime.datetime.now() has 2 more bytes than __import__('datetime').datetime.now(). Emacs Lisp, 83 bytes (run-with-timer 1 1(lambda()(message(format-time-string"%H:%M:%S"(current-time)))))  Groovy, 40 37 bytes Not very golfable :(, thanks to FlagAsSpam for for(;;) :) for(;;){print(new Date());sleep 1000}  • for(;;){...? And what about 1e3 for 1000? You can use print instead of println. – Addison Crump Dec 16 '15 at 14:30 • Thanks, I was pretty sure it needs to be a new line every time. – Fels Dec 16 '15 at 14:36 bash, 35 bytes Just for kicks yes|mapfile -tc1 -C'date;sleep 1 #'  Also 35, x='x[date>&2;sleep 1${!x}]'<${!x}  I think it's unlikely there's anything shorter than exec$0.

• Are you sure the first solution is not als0 35 characters long? I have to put a space in front of the # otherwise sleep cries for invalid syntax. – manatwork Dec 16 '15 at 14:26
• @manatwork Yes. Looks like I tested a version that did ;: before I changed it to # and made a typo. The interactive_comments shopt sometimes screws with me too. Good catch. – ormaaj Dec 16 '15 at 15:11

Python 3, 48 bytes

from time import*
while[sleep(1)]:print(ctime())

It is a @Mego's Python 2 solution tweaked to work on Python 3.
Here's Jupyter notebook to see the results and try it.

Lua, 56 bytes

Lua is pretty hard to golf...

v=1
while 1 do v=os.date'%c' l=v~=l and print(v)or v end

There are few online interpreters that allow infinite loops; one of the few I've found is this one. This repeatedly converts the current system time to a date-time string and outputs that string when it's different from the last.

Ungolfed:

current = 1
while 1 do
current = os.date('%c')
last = current ~= last and print(current) or current
end

• +1 for golfing in lua :). You can golf it down to 51 bytes by using a label+goto to do an infinite loop, and suppress some spaces : v=1::a::v=os.date'%c'l=v~=l and print(v)or v goto a – Katenkyo May 23 '16 at 6:49

Matlab, 50 42 bytes

while 1;pause(1);disp(datestr(now,13));end

• I believe you can use the predefined numeric date format 13 instead of 'HH:MM:SS' and save 8 bytes. – beaker Nov 30 '15 at 17:54
• Somehow I'm noticed this a bit late, but thank you very much! – flawr Jan 24 '16 at 11:18
• Perfect score! :D – beaker Jan 24 '16 at 15:02

Perl 6, 34 30 bytes

Perl 6 is looooong, but I like it. Yes, all the whitespace is needed.

sleep 1 while say DateTime.now

Hooray for postfix syntax!

F#, 101 98 bytes

open System;while 1=1 do(DateTime.Now.ToString"HH:MM:ss"|>printfn"%s";Threading.Thread.Sleep 1000)

Pretty straightforward: print the current time in the correct format, wait a second, then repeat. Forever.

Credit to @VoteToClose and @RikerW for the help.

• Few questions: is for(;;) valid syntax? Can you use DateTime.Now.ToString"HH:MM:ss"? Do you need the whitespace in ; Threading.Thread.Sleep – Addison Crump Feb 2 '16 at 21:42
• Few answers! From what I've seen, for(;;) sadly isn't valid F#. The other two suggestions are dead on, however - I can't believe I've overlooked those. Thanks! =) – Roujo Feb 2 '16 at 21:46
• while (truthy value)? – Rɪᴋᴇʀ Feb 2 '16 at 21:48
• @RikerW Right, that's another good one. I usually write infinite loops as recursive functions, so I hadn't thought about it. Thanks! =) – Roujo Feb 2 '16 at 21:51
• Not just plain 1? – Rɪᴋᴇʀ Feb 2 '16 at 21:58

Rebol, 25 bytes

forever[print now wait 1]

This prints day & time (24 hour HH:MM:SS). For an extra 5 bytes you can make it print just the time: forever[print now/time wait 1]

• Lol, is it said that I read this like house music? Print, now wait 1. Print, now wait 1. – Addison Crump May 20 '16 at 9:35


text:
# Find the equation of a tangent line at each given point

in title

## Homework Equations

x=2cotθ
y=2sin2θ
dy/dx = 2sin3θcosθ
y-y1=m(x-x1)
point = (-2/√3,(3/2)

## The Attempt at a Solution

Have been stuck for hours

I solved for the dy/dx above, now I need to figure out how to get rid of the θ to get my equation in terms of x
so I was thinking sqrt(y/2) = sinθ and (x/2)*sinθ = cosθ

is this a correct approach?

## Answers and Replies

Apparently ##\theta## is a parameter: ##x=x(\theta), y=y(\theta)##. Is that right? So you can write ##sin(\theta)=f(y), cos(\theta)=g(y)##. Then you eliminate ##\theta## and get ##x=x(y)## and ##\frac{dx}{dy}=h(y)##.

isukatphysics69
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
There is no need to get rid of ##\theta## in order to figure out the slope. What you do need to do is to figure out what value of ##\theta## you should be using, i.e., which value of ##\theta## corresponds to the given point.

isukatphysics69
@Orodruin is right; I don't know what I was thinking of.

isukatphysics69
How is the point point = (-2/sqrt(3),(3/2) the same as 2pi/3?


text:
# Tensile Force on Line of Bolts

I'm presented with a problem requiring some analysis to determine if an antenna bracket mounted to a concrete blockwork wall is sufficiently strong enough to endure a wind load given the known surface area of two antennas mounted on it.

Excuse the terrible diagram:

The force due to the surface area of the antennas & wind velocity is assumed to be 288N at the end of the vertical post acting out of the page and attempting to 'lever' bolts B1:B3 out of the wall (pullout/tensile force).

Distance B1 to force = 1250mm, B2 to force = 912mm, B3 to force = 742mm

I've simplified the problem to assume B2 doesn't exist, B1 is the fulcrum and B3 is the 'load' in a class 2 lever system:

For the simplified system I work out the tensile force acting on B3 to be in the realm of 709N:

(1250/(1250-742))*288N = 708.67N

Having an electrical background I'd be very interested to hear how a mechanical engineer would approach the problem and whether there is a better way (than my simplified system) to represent the forces on each of the 3x bolts.

## 3 Answers

For small equipment mounting, let me introduce two simple methods:

1) Rigid Rotation Method (Classic)

2) Cantilever Method (Conservative approximation)

• Assume cantilever action, for which the applied load rotates about the point "A", and the rotation is restricted by the two bolts farthest from the point "A".

Conclusion:

I recommend using the Metod 2 for its simplicity and embed safety. Also, I suggest adding a rod from the fixture to the wall to take care of the vibration induced by wind.

Please check my math for mistakes.

Assuming the lever is rigid enough not to bend to simlify our estimate:

The lever will rotate about B1 bolt because it can not penetrate the concrete block. So B2 to and B3 will share the 288N proprtionaly to their distance from B1. Lets call their reactio RB2 and RB3 respectively.

$$R_{B3} =(1250-742=508)/(1250-912=338)*R_{B2}= 1.5 R_{B2}$$

By substituting tension in B3 as 1.5 B2 we get

$$\Sigma M_{B1}= 0, \ R_{B2}*338mm+ 1.5 R_{B2}*508mm= 288N*1250 \ \rightarrow \\ R_{B2}=327N \ R_{B3} =327*1.5 = 490N$$

As mentioned in kamran's answer its very important that the bar is rigid. Otherwise the following analysis is not relevant.

Although you don't seem to consider it this way, -if the beam is rigid enough-, then the fulcrum at the far left edge of the beam (not on B1, so the force there is not zero). The equation I derived is based on that assumption, which is more correct and will give slightly different results (the accuracy will depend on the beam's rigidity).

Below is an image of the side and top view of the beam (bracket to my mind has a right angle somewhere)

The top view (its at the bottom - I'm using a first angle projection), shows

• the undeformed with solid lines and
• the deformed position with light blue and dashed lines.

From there you can see that the displacement for each bolt, is going to be proportional to the distance of the bolt from the fulcrum. The fulcrum is the far left of the beam. Because the force is proportional to the displacement then the force on each bolt will be equal to:

$$F_i = K\cdot L_i$$ where:

• $$F_i$$ is the force on each bolt
• $$K$$ is the spring coefficient of the bolt (which eventually you need to calculate)
• $$L_i$$ is the distance of the bolt from the fulcrum .

Then you can write the equations for the equilibrium of moments about the fulcrum :

$$F_1\cdot L_1 + F_2\cdot L_2+ F_3\cdot L_3 - F\cdot L_F =0$$ or $$\left(K L_1\right)\cdot L_1 + \left(K L_2\right)\cdot L_2+ \left(K L_3\right)\cdot L_3 - F\cdot L_F =0$$ if you simplify that: $$K \left(L_1^2 + L_2^2 + L_3^2\right)=F\cdot L_F$$ therefore: $$K =F\cdot \frac{L_F}{\left(L_1^2 + L_2^2 + L_3^2\right)} = \frac{L_F}{\sum_{i=1}^{3}L_i^2} \cdot F$$

Now that you have K, you can calculate : $$F_i=K \cdot L_i$$

## numerical example

This is to illustrate the difference, of selecting a different position to the fulcrum point.

if you assume the fulcrum is at B1, then you end up with (essentially the same with kamran's computation)

$$F_1=0 N, \qquad F_2= 326.86 N , \qquad F_3= 491 N$$

if you assume that the fulcrum is 100 mm to the left then the above values become $$F_1=68 N, \qquad F_2= 297 N , \qquad F_3= 413 N$$


text:
# Correctness proof: induction on sequence of steps, need a stronger claim?

Im trying to prove the correctness of the construction proposed in this site answer: a two stack PDA that simulates a Turing Machine. By "correctness" i mean to prove more or less formally that we can translate accepting sequences of steps.

Given a TM $$M=(Q,\Sigma_I,\Sigma_O,\delta,q_0,Q_F)$$ and a 2-PDA $$A=(Q',\Sigma_I,\Sigma_O',\delta',q_0', Q_F)$$ defined as the cited answer suggest.

We need to show that for all $$w \in \Sigma^*$$, if $$w \in L(M)$$ then $$w \in L(A)$$, i.e, if the first accepts then the other accepts.

This is for all $$w \in \Sigma^*$$: $$\exists q \in Q_F: q_0\,w \Rightarrow_{\delta}^* \alpha_1\,q\, \alpha_2 \,\,\implies\,\, \exists q' \in Q_F: (q_0',w,\,\) \Rightarrow_{\delta'}^* (q',\epsilon,\beta_1,\beta_2) \tag{*}\label{*}$$ where:

• $$\Rightarrow_{\delta}$$ is the usual step relation between machine descriptions, asume it is defined appropriately for TM and for 2-PDA based on the transition functions $$\delta$$ and $$\delta'$$ .
• $$\Rightarrow_{\delta}^*$$ is the transitive-reflexive closure of step relation, also for TM and 2-PDA.
• Descriptions for TM are noted $$X_1...X_{i-1} q X_{i}... X_n$$ meaning the actual tape content is $$X_1...X_n$$, actual state is $$q$$ and head is over $$X_i$$
• Descriptions for 2-P2A are quadruplets $$(q,w,\beta_1,\beta_2)$$ meaning the actual state is $$q$$, remaining input is $$w$$, first stack content $$\beta_1$$ and second stack content $$\beta_2$$.

So $$\eqref{*}$$ is saying that for any word $$w$$, if there is in $$M$$ a sequence of steps in from $$q_0$$ to a final state $$q$$ then there is a sequence of steps in $$M'$$ from $$q_0'$$ to a final state $$q'$$.

Induction over the length of $$\Rightarrow_{\delta}^*$$ would be sketched as:

1. Base case $$n=1$$ .....
2. Induction Hypothesis: Asume for any sequence of length $$n > 0$$.
3. Induction Thesis: Prove for a sequence of length $$n+1$$ where $$n > 0$$ .....

But... i have a problem with this approach. In the proof for the inductive step, we have a sequence of $$n+1$$ steps for $$M$$ like: $$q_0\,w \,\Rightarrow_{\delta}\, ... \Rightarrow_{\delta}^n\, \gamma_1\,q_i\,\gamma_1 \,\Rightarrow_{\delta}\, \alpha_1\,q\,\alpha_2 \quad\text{ and } q \in Q_F$$

I cant apply here the Inductive Hypothesis because $$q_i$$ cant be a also a final state transitioning to $$q$$. Final states in TM can't have ongoing transitions ( maybe relaxed variants can be proposed but i prefer to let it be like that).

My question is, do i need to prove something stronger than $$\eqref{*}$$ to avoid these little technical inconveniences?. Im thinking something like: $$\forall q \in Q: q_0\,w \Rightarrow_{\delta}^* \alpha_1\,q\, \alpha_2 \,\,\implies\,\, \forall q' \in Q': (q_0',w,\,\) \Rightarrow_{\delta'}^* (q',\epsilon,\beta_1,\beta_2)$$ But im not sure this logically implies $$\eqref{*}$$. I would like to hear general recommendations when doing this type of proofs.


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## 6166: The Black Hole of Numbers (20)

[提交][状态][讨论版][命题人:]

## 题目描述

For any 4-digit integer except the ones with all the digits being the same, if we sort the digits in non-increasing order first, and then in non-decreasing order, a new number can be obtained by taking the second number from the first one. Repeat in this manner we will soon end up at the number 6174 -- the "black hole" of 4-digit numbers. This number is named Kaprekar Constant.

For example, start from 6767, we'll get:

7766 - 6677 = 1089
9810 - 0189 = 9621
9621 - 1269 = 8352
8532 - 2358 = 6174
7641 - 1467 = 6174
... ...

Given any 4-digit number, you are supposed to illustrate the way it gets into the black hole.

## 输入

Each input file contains one test case which gives a positive integer N in the range (0, 10000).

## 输出

If all the 4 digits of N are the same, print in one line the equation "N - N = 0000". Else print each step of calculation in a line until 6174 comes out as the difference. All the numbers must be printed as 4-digit numbers.

## 样例输入

6767
2222

## 样例输出

7766 - 6677 = 1089
9810 - 0189 = 9621
9621 - 1269 = 8352
8532 - 2358 = 6174
2222 - 2222 = 0000

[提交][状态]


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# Demonstration of the differentiability of a continuous function

1. Sep 29, 2010

### Telemachus

1. The problem statement, all variables and given/known data
I have some doubts about the demonstration of the differentiability. If I'm asked to proof that an average function is differentiable on all of it domain, lets suppose its a continuous function on all of its domain, but it has not continuous partial derivatives. How should I demonstrate that its differentiable? May I use the limit with generic points $$(x_0,y_0)$$? I mean, if I use this limit (the one with the function and the tangent plane over the square root that represents a disk), and its a differentiable function, with this generic points the limit should give zero, right?

Bye there, thanks for posting.

PD: I'm talking for function of two or more variables.


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+0

# Stats help thank you

0
124
2
+125

a high school  basketball players point totals for the regular season are normally distributed with a mean of 17 and a standard deviation of 4

sarah has a point z-score of 2.8

what is sarahs point total for the regular season? And does this point total seem unusual? Explain

Mar 27, 2019

#1
+5662
+1

$$T = \mu + z \sigma = 17 + (2.8)(4) = 28.2\\ P[\text{someone scores > Sarah}] = 0.00255513\\ \text{So Sarah's score is pretty exceptional}$$

.
Mar 27, 2019
#2
+1

Using a positive z-score table reveals  z = 2.8  is a score better than 99.74% of other players !

2.8 x 4  + 17 = 28.2 pts total    for Sarah....

Mar 27, 2019


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# Cauchy real numbers with and without modulus

In constructive mathematics there are many possible inequivalent definitions of real numbers. The greatest variety seems to be in Dedekind-style approaches: in addition to "the" Dedekind real numbers which satisfy locatedness (for all $q<r$ in $\mathbb{Q}$, either $x<r$ or $q<x$), there are the MacNeille, upper, and lower reals, and perhaps others. It is easy to construct models distinguishing between all these: for instance, in a topos of sheaves on a space $X$, the Dedekind reals are the sheaf of continuous $\mathbb{R}$-valued functions on $X$, the upper and lower reals are respectively the sheaves of upper and lower semicontinuous $\mathbb{R}$-valued functions on $X$, and the MacNeille reals are the sheaf of pairs of upper and lower semicontinuous functions satisfying a closeness condition. This gives me a good topological intuition for the difference between these definitions.

This question is about an analogous possible variation in Cauchy-like real numbers. Classically, a sequence $(x_n)$ of rational numbers is Cauchy if $\forall k \exists n \forall p,q>n. |x_p-x_q|<2^{-k}$. But almost without exception, constructive mathematicians define Cauchy sequences by "building a Skolem function" into the definition, assuming that there is a "modulus of Cauchy-ness" $N:\mathbb{N}^{\mathbb{N}}$ such that $\forall k \forall p,q > N_k. |x_p-x_q|<2^{-k}$. And once one has such a modulus, one can massage the sequence $(x_n)$ to make the modulus coincide with a standard one such as $N_k = k$.

My question is: what is the difference in constructive mathematics between Cauchy real numbers having a modulus of Cauchy-ness and without it? It seems to me that the no-modulus Cauchy reals sit in between the with-modulus Cauchy reals and the Dedekind reals; we shoudn't need a modulus to get locatedness, all we need is the existence of some point of the sequence that's closer than $\frac{|q-r|}{2}$ to the limit so we can compare it to $q$ and $r$.

I understand that someone who cares about algorithms and computation will want a modulus to compute with, but if I am a topologist and am happy with Dedekind reals, leaving out a modulus in the Cauchy reals is not a priori a bad thing to do. Are the no-modulus Cauchy reals any less well-behaved intrinsically than the with-modulus ones? Is their algebraic and order structure any different? And are there (hopefully topological) models (necessarily failing countable choice) that distinguish the no-modulus Cauchy reals from both the with-modulus ones and from the Dedekind reals?

• If I remember correctly, the problem with Cauchy real without modulus is that in general you can't apply the usual diagonal process showing that a Cauchy sequence of Cauchy real converge to a Cauchy real... This makes their study quite limited. Also it is indeed true that any Cauchy real (with or without modulus) is a Dedeking real. – Simon Henry Jan 4 '18 at 16:53
• @SimonHenry You can't do that for Cauchy reals with modulus either unless you assume countable choice (thereby collapsing the Dedekind reals to the Cauchy reals as well), since a Cauchy sequence of Cauchy reals is a sequence of equivalence classes of Cauchy sequences. – Mike Shulman Jan 4 '18 at 17:24
• I don't follow your locatedness argument: how do you know when you have a term within $\varepsilon$ of the limit? – François G. Dorais Jan 4 '18 at 17:25
• Just to clarify the terminology: am I understanding right that by “with-modulus”, you don’t mean “equipped with a modulus”, but rather “such that a modulus exists”? – Peter LeFanu Lumsdaine Jan 4 '18 at 17:43
• @PeterLeFanuLumsdaine Haha, good question! Certainly a given Cauchy sequence should represent the same real number no matter which modulus you give it. We're already passing to equivalence classes of Cauchy sequences in defining the Cauchy reals, so the question is about comparing equivalence classes of Cauchy sequences for which there exists a modulus to equivalence classes of Cauchy sequences equipped with a modulus. But I think these are isomorphic, as long as we have the function comprehension (unique choice) principle. – Mike Shulman Jan 4 '18 at 17:49

I don't see an algebraic difference. Even in constructive math without countable choice, we can define the total functions $+, -, \times, \min, \max, \sqrt[3]{}$ and the partial functions $1/x, \sqrt{x}$ on the no-modulus Cauchy reals.

However, there is a difference in that Cauchy completeness does not hold for the no-modulus reals. Consider the situation in the recursive topos or in recursive mathematics:

• Let $R$ be the no-modulus Cauchy reals, considered as functions $r : N \rightarrow Q$.
• Let $S$ be the set of with-modulus Cauchy sequences of no-modulus reals, considered as functions $s : N \rightarrow R$, such that $\forall m, m', |s(m)-s(m')| < 1/m + 1/m'.$
• Suppose (for contradiction) that $\forall s \in S\ \exists r \in R\ \lim s=r$.
• Then there would be a recursive function $f:S\rightarrow R \ \lim s=f(s)$.
• Now consider $s=0^S$, i.e. the sequence such that $\forall n\ s(n)=0^R$, i.e. the sequence such that $\forall n\forall m\ s(n)(m)=0^Q$. The calculation of $(f(s))(1)$ uses at most $k_1$ terms of $s$, the calculation of $(f(s))(2)$ uses at most $k_2$ terms of $s$, etc.
• Let $t(n)(m)=0$ if $n+m < k_{n+m}$, and 1 otherwise. Then $t$ is in $S$.
• Then $f(s)=f(t)$, but $\lim s = 0$ and $\lim t=1$, which is a contradiction.

If we tried to replace $R$ and $S$ by the with-modulus Cauchy reals and sequences thereof, we would find that $t(1)$ might not be a with-modulus real, $t$ might not be in the new $S$, and the argument would not go through.

So in the recursive world: the with-modulus reals lead to all Cauchy sequences converging; the no-modulus reals lead to Cauchy sequences that may not converge.

• This is an interesting argument. In general I don't know of any way to prove that the with-modulus Cauchy reals are Cauchy complete without using countable choice, which already collapses the Dedekind reals to the with-modulus Cauchy reals. You seem to be saying there is a particular model in which the with-modulus Cauchy reals don't coincide with the Dedekind reals, but happen to nevertheless be Cauchy complete, while the no-modulus ones aren't; is that right? – Mike Shulman Jan 4 '18 at 17:26
• Is there a logical principle that implies Cauchy completeness of the with-modulus Cauchy reals but is weak enough to keep them distinct from the no-modulus ones? – Mike Shulman Jan 4 '18 at 17:27
• @MikeShulman, when you said "almost without exception, constructive mathematicians...", I think it was fair to assume no equivalence classes. So perhaps you might post a separate question, like this: Let's work in the context of constructive mathematics without countable choice. Consider E = Cauchy sequences of rationals with explicit modulus and F = Cauchy sequences of rationals with free modulus. Consider the corresponding sets of equivalence classes E* and F*. Obviously there is no canonical isomorphism between $(E^*,+,*,<)$ and $(F^*,+,*,<)$. But what properties distinguish the two? – Matt F. Jan 4 '18 at 20:55
• For more on the subject of properties of Cauchy sequences (as opposed to reals), see doi.org/10.1016/j.entcs.2006.09.012 – Dap Jan 5 '18 at 6:37
• @Dap Thank you!! Could you please post that paper in an answer to the question, not just in a comment? It doesn't exactly address the question I asked, but it's highly relevant and very comprehensive in what it does address. – Mike Shulman Jan 8 '18 at 19:12

Let $(a_n)$ be a Specker sequence - a computable, bounded, increasing sequence of rationals so that the limit is not a computable real.

First, assume for simplicity that we work in any system of second-order arithmetic (classical or not) that has REC as a model. This is the model with the standard natural numbers and the computable sets of natural numbers. In this model, the Specker sequence is a non-modulus Cauchy sequence (because it is a Cauchy sequence in the standard model, and being Cauchy is defined arithmetically). But there is no computable modulus of convergence for a Specker sequence, so the model does not believe that the sequence is a modulus Cauchy sequence.

Thus our system cannot prove that "every non-modulus Cauchy sequence has a modulus". Systems affected by this include many systems of constructive second-order arithmetic as well as the classical system $\mathsf{RCA}_0$.

In fact, every modulus of a Specker sequence computes $\emptyset'$. So the previous argument also goes through for any system of second-order arithmetic (classical or not) that has a $\omega$-model without any Turing complete reals. This applies to $\mathsf{WKL}_0$ and to constructive systems with various forms of compactness.

Similar arguments will work for systems that are not fragments of second-order arithmetic, as long as they have $\omega$-models that do not contain any Turing complete reals.

• Thanks! I find it easier to think about toposes, but presumably there is a realizability topos containing a Specker sequence in which the same argument works. But what I would really understand much better is a topological model. – Mike Shulman Jan 4 '18 at 17:39
• Hmm... I don't know much about recursive models, but I'm looking at the construction of a Specker sequence in Troelstra and van Dalen, and it looks to me as though it proves that if $f:\mathbb{N}\to\mathbb{N}$ is injective and $r_k = \sum_{i=0}^k 2^{-f(i)}$, and if $(r_k)$ is no-modulus Cauchy, then the image of $f$ is decidable -- from which I think one can deduce that $(r_k)$ in fact converges and is with-modulus Cauchy. So I don't understand how a Specker sequence can be no-modulus Cauchy. – Mike Shulman Jan 4 '18 at 19:53
• @Mike Shulman: it's a bounded increasing sequence of rationals, so classically it is a Cauchy sequence (ignoring the issue of a modulus). Because the sentence expressing that this Specker sequence is (no-modulus) Cauchy is arithmetical, and because the definition of the sequence itself is arithmetical, this means that every $\omega$-model will believe that the sequence is no-modulus Cauchy. – Carl Mummert Jan 4 '18 at 21:53
• Looking at Troelstra and van Dalen p. 268, to show that the image of $f$ is computable using their construction requires a modulus. They say "then for any $m$ we can find a $k$ with ..." - but in order to compute the range of $f$ we would need to compute $k$ from $m$, which is to say we need a computable modulus. (They are also working in the context of $\text{CT}_0$ there, but this is primarily to bridge the gap between the constructive meaning of "decidable" and the computable meaning. The range of $f$ does not exist in REC, which does not satisfy $\text{CT}_0$) – Carl Mummert Jan 4 '18 at 22:14
• Indeed, they comment on p. 193 that $\text{CT}_0$ is equivalent to saying that the function implicitly defined by an arithmetical $(\forall n)(\exists m)$ statement is always computable; finding the modulus of a Specker sequence is just a special case of this. – Carl Mummert Jan 4 '18 at 22:40


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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

Many statistical learning algorithms perform better when the covariates are on similar scales. For example, it is common practice to standardize the features used by an artificial neural network so that the gradient of its objective function doesn’t depend on the physical units in which the features are described.

The same advice is frequently given for K-means clustering, but there’s a great counter-example given in The Elements of Statistical Learning that I try to reproduce here.

Consider two point clouds ($n=100$ each), randomly drawn around two origins 3 units away from the origin:

set.seed(495)
n <- 100
d <- 3
x <- matrix(rnorm(n * 2, sd = 1), ncol = 2)
x[1:(n/2), 1] <- x[1:(n/2), 1] - d
x[(n/2 + 1):n, 1] <- x[(n/2 + 1):n, 1] + d

The K-means algorithm has no problem in classifying these points:

km <- kmeans(x, centers = 2)
km\$centers

##        [,1]         [,2]
## 1  2.922143  0.098422541
## 2 -2.991026 -0.003131757

Let’s see now what happens when we standardize each feature. Since their mean is already zero, we merely divide by their standard deviation:

x_scaled <- x
x_scaled[, 1] <- x_scaled[, 1] / sd(x_scaled[, 1])
x_scaled[, 2] <- x_scaled[, 2] / sd(x_scaled[, 2])

And we run again the K-means algorithm on these new data:

km_scaled <- kmeans(x_scaled, centers = 2)

We see that K-means has completely failed to identify the clusters, because ‘standardizing’ the features has destroyed the clear separation between the clusters.

So what’s the lesson here? Clearly, for K-means you should not blindly standardize the features unless there are clear reasons to do so. In this toy example, we didn’t know what the features represent, so it’s impossible to say whether standardizing the features was the right thing to do. Perhaps the clusters seen pre-standardization were mere artefacts of our choice of units! As a rule of thumb, I suggest that features that are expressed in the same units and that represent the same ‘stuff’ (such as width and length) should not be standardized. If you have deeper insights into this I’d love to hear your comments.

The post Feature standardization considered harmful appeared first on David's blog.


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4.1. TEST YOURSELF

Generating Candidate Lists of Adaptive TEs

1.
How many TEs are in genome regions with recombination (non-zero)?

2. Remember, once you take a subgroup from the original file, the following subsets will not be taken from the original file but from the previous subgroup you obtained.

How many TEs are in recombination regions and in high frequency regions (more than 10%) in the genome of at least one out of the 5 populations?

3. We want to detect the adaptive “out-of-Africa” TEs, meaning these flies originated under tropical African conditions and then had to adapt to the cold and arid conditions of Europe. In this case, which populations will we will have to compare?

4. Keep in mind that with R, we can get a subset from the original file, taking into account more than one condition:

new_list <- subset (last_list_data_file, condition_variable | condition_variable & condition_variable

You can check it in the code file opened in the text editor. The threshold in which we consider a TE low frequency or  high frequency is a hot point. We consider a TE low frequency if less than 10% of the population carry it; and high frequency if more than 10% of the population carry it. By using the subset function, you can generate a list of which “out-of-Africa” candidates are adaptive. In this case, how many candidate TEs do we have for the “out-of-Africa” adaptation? (Remember they should be in regions with recombination).

5. We consider those TEs located inside genes or in the nearby gene regulatory regions to be most important. This way, TEs in our lists will be more likely to affect the nearby gene and avoid including neutral TEs. In Drosophila melanogaster, promotor and regulation regions are considered to be 1kb from the gene.

Remember that in our data file, gene distance is measured in bp and not in kb. And remember, module 2 box to transform the number and get the information you want.

How many candidate TEs do we have for the “out-of-Africa” adaptation (regions with recombination), within less than 1kb of a gene?


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# The ratio of actual velocity to the local velocity of sound is called:

Free Practice With Testbook Mock Tests

## Options:

1. Mach number

2. Velocity ratio

3. Velocity factor

4. Speed ratio

### Correct Answer: Option 1 (Solution Below)

This question was previously asked in

RSMSSB JE Mechanical (Degree) held on 13/12/2020

## Solution:

Explanation:

The following table explains the difference between the four options.

Difference between Mach number, velocity ratio, and factor, speed ratio respectively.

 S. No. Ratio/factor Definition Formula Significance Mach Number It is the ratio of flow velocity c past a boundary to the local speed of sound v. It is a dimensionless quantity. $$M = \frac{c}{v}$$ Used in Fluid mechanics The nature of propagation of disturbance depends on Mach number. Velocity Ratio It is the ratio of the angular velocity of the driven wheel $${\omega _{driven}}$$ or gear to the angular velocity of driver wheel or pinion $${\omega _{driver}}$$. It is a dimensionless quantity. $$V = \frac{{{\omega _{driven}}}}{{{\omega _{driver}}}}$$ Used in Theory of machines. Gear analysis Belt power transmission analysis. Velocity factor It is the ratio of the speed at which a wavefront (of an electromagnetic signal passes through the medium, to the speed of light in vacuum. The formula depends on the application. For optical signals $$VF = \frac{1}{n}$$ n is the refractive index of the medium. Used majorly in optics. Speed Ratio It is the ratio of the speed of the driving shaft to the driven shaft. It can be positive or negative, unlike the velocity ratio. $$SR = \frac{{{N_{driver}}}}{{{N_{driven}}}}$$ Used in Theory of machines. Gear train analysis. It is reciprocal of the train value.

 Mach number M Type of flow M > 1 Sub-sonic flow M < 1 Super-sonic flow M = 1 Sonic flow


text:
HAL will be down for maintenance from Friday, June 10 at 4pm through Monday, June 13 at 9am. More information

# Meta-Theoretical Properties of lambda_phi: A Left-Linear Variant of lambda_sigma

Abstract : In this paper we consider explicit substitutions calculi that allow open terms. In particular, we propose a variant of the $\lambda_\sigma$-calculus, that we call $\lambda_\phi$. For this calculus and its simply-typed version, we study its meta-theoretical properties. The $\lambda_\phi$-calculus enjoys the same general characteristics as $\lambda_\sigma$, i.e. a simple and finitary first-order presentation, confluent on terms with meta-variables, with a composition operator and with simultaneous substitutions. However, $\lambda_\phi$ does not have the non-left-linear surjective pairing rule of $\lambda_\sigma$ which raises technical problems in some frameworks.
Document type :
Reports
Domain :

https://hal.inria.fr/inria-00073584
Contributor : Rapport de Recherche Inria Connect in order to contact the contributor
Submitted on : Wednesday, May 24, 2006 - 1:14:51 PM
Last modification on : Thursday, February 3, 2022 - 11:18:27 AM
Long-term archiving on: : Sunday, April 4, 2010 - 11:50:54 PM

### Identifiers

• HAL Id : inria-00073584, version 1

### Citation

César Muñoz. Meta-Theoretical Properties of lambda_phi: A Left-Linear Variant of lambda_sigma. [Research Report] RR-3107, INRIA. 1997. ⟨inria-00073584⟩

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# The stellar orbit distribution in present-day galaxies inferred from the CALIFA survey

Zhu, L, Ven, GVD, Bosch, RVD, Rix, H-W, Lyubenova, M, Falcón-Barroso, J, Martig, M, Mao, S, Xu, D, Jin, Y, Obreja, A, Grand, RJJ, Dutton, AA, Maccio, AV, Gómez, FA, Walcher, JC, García-Benito, R, Zibetti, S and Sánchez, SF (2018) The stellar orbit distribution in present-day galaxies inferred from the CALIFA survey. Nature Astronomy. ISSN 2397-3366

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Text
1711.06728v3.pdf - Accepted Version

Galaxy formation entails the hierarchical assembly of mass, along with the condensation of baryons and the ensuing, self-regulating star formation. The stars form a collisionless system whose orbit distribution retains dynamical memory that can constrain a galaxy's formation history. The ordered-rotation dominated orbits with near maximum circularity $\lambda_z \simeq1$ and the random-motion dominated orbits with low circularity $\lambda_z \simeq0$ are called kinematically cold and kinematically hot, respectively. The fraction of stars on cold' orbits, compared to the fraction of stars on hot' orbits, speaks directly to the quiescence or violence of the galaxies' formation histories. Here we present such orbit distributions, derived from stellar kinematic maps via orbit-based modelling for a well defined, large sample of 300 nearby galaxies. The sample, drawn from the CALIFA survey, includes the main morphological galaxy types and spans the total stellar mass range from $10^{8.7}$ to $10^{11.9}$ solar masses. Our analysis derives the orbit-circularity distribution as a function of galaxy mass, $p(\lambda_z~|~M_\star)$, and its volume-averaged total distribution, $p(\lambda_z)$. We find that across most of the considered mass range and across morphological types, there are more stars on warm' orbits defined as $0.25\le \lambda_z \le 0.8$ than on either cold' or `hot' orbits. This orbit-based "Hubble diagram" provides a benchmark for galaxy formation simulations in a cosmological context.


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# Relativity in a current carrying wire

I thought I had the conceptual understanding of how magnetism between a charged particle and a current carrying wire can be seen from the point of view of relativistic effects, but after revisiting the concept, I got conceptually stuck.

The easiest way to show my issue is to reference this video (between 1:20 and 2:30): https://www.youtube.com/watch?v=1TKSfAkWWN0

THE ARGUMENT:

The argument goes that in the reference frame shared by us and the stationary particle (stationary with respect to the wire and the positive charges in it), the positive charges in the wire are stationary, while the negative charges drift along with some velocity. The claim is that the two types of charge have similar charge densities, so the attraction and repulsion effects cancel, and we perceive no force on the test particle.

Next we have the test particle start moving with the same velocity v, so that now it's the negative charges that look stationary, and the positive charges are "moving". Because of relativstic effects of length contraction, the positive charges seem to get closer, while the negative charges spread out to their "true" separation as seen from their reference frame. The net effect is that the positive charge density is now greater, so the test particle will feel a force.

MY CONFUSION:

The story makes sense expect the first scenario has the negative charges moving with respect to the "stationary" test particle. So these charges need to have their length contracted relative to what we'd see if they were not moving relative to the test particle/charge.

!!! So my central issue is with the claim at the beginning that the positive and negative charge densities are the same and would cancel for the "stationary" particle. !!!

The way I see it, the test particle will see the moving negative charges as contracted and the positive stationary ones as not contracted, resulting in a higher charge density for the negative current. That forces me to conclude the positive test charge will then be attracted toward the wire.

I END UP WITH THREE SCENARIOS (Which I know are wrong):

First, we have a wire with positive charges and negative charges, and a test particle/charge. There's no current yet, so the positives and negatives in the wire are not contracted, and so have the same charge density. Nothing happens. Then we flip a switch and cause a current to flow.

• Test particle is stationary, sees stationary positive charges and moving + contracted negative charges. The negative charge density is greater, so the test charge feels a force toward the wire.
• Test particle moves in the direction of the negative current, but with half the velocity of the current. This setup results in the test particle seeing both the negative and positive charges in the wire appearing to move away from the test particle with the same velocity but opposite directions. So both would be equally contracted in distance, resulting in equal charge densities, leading to no force on the test particle.
• Test particle moves along at the same velocity as the negative charge, so this charge is not contracted, but now the positive charges in the wire are contracted and as much so as the negative charges were in the first scenario. So now the positive charge density is greater, and the particle feels a force away from the wire.

This is obviously NOT what occurs in reality, so I've reasoned myself into a corner...

Thank you for any suggestions!

• I'm interested in writing an answer, however I want to clarify something about your 3 scenarios. These are to be different physical situations that are examined in different reference frames. (Example: in scenario #1, the test charge is in the same frame as the positive charges, but in scenario #3, the test charge is in the same frame as the negative charges.) Because they are different scenarios the physical results will be different. Are you sure these are the scenarios you want examined? Usually in relativity you look at the same situation from different frames, so I want to make sure. – eigenchris Jul 29 '20 at 17:58
• @eigenchris Thank you for your interest! I made the scenarios from thinking: "ok, if I kept going with my wrong understanding, what setups could I make so I get an attractive/repulsive/zero force effect in the system". I think my main trouble is with the fact that we see the negative line of current flowing and length contracted, yet its density balances with the density of the stationary and not contracted positive line of charges. To me, this implies stopping the current will expand the charges out, decreasing the - density relative to the + density, implying a neutral wire isn't neutral... – amazonprime Jul 29 '20 at 18:55
• You are exactly correct that stopping the current will cause the negative charge to spread out, resulting in an overall positive net charge... and the wire isn't neutral. My biggest problem with that segment between 1:20 and 2:30 is that the presenter starts with an overall neutral wire... but the instant the negative charges start moving (at 1:28), he has actually changed the physical situation entirely, and switched to a wire that is neutral only when negative charges are moving, but postive overall if the current stops. The wires before 1:28 and after 1:28 are completely different. – eigenchris Jul 29 '20 at 19:12
• @eigenchris Ah, I think that really helps! So basically, if I understand well, there's no contradiction in a wire being neutral both when it has current and when it doesn't, because the relativistic effects aren't the only thing in play and you'd have to also account for how the wire changed from one state to the other. – amazonprime Jul 29 '20 at 19:34
• I think I agree with that. I can write up a full answer later today. – eigenchris Jul 29 '20 at 19:55

The circuit is presumably built from electrically neutral wire, containing the same number of positive and negative charges. When the circuit is switched on and current starts to flow, new negative charges can't appear from nowhere. The average spacing of charges (measured in the lab frame) is the same before and after the current is switched on, because it's equal to the total length of the wire (measured in the lab frame) divided by the number of charges, neither of which changes.

There will be transient pressure waves of more and less densely packed charges when the circuit is switched on, and even in the steady state there may be a slight variation in the density of charges around the circuit, but those effects have nothing to do with length contraction. They would exist in a hydraulic circuit in a Newtonian universe too.

With respect to an inertial frame moving relative to the lab frame, there will be a nonzero charge density in parts of the wire, but the total charge density summed/integrated over the whole wire will still be zero because charge is a relativistic invariant.

You could hypothetically build a circuit from a wire with a factor of $$γ$$ more negative than positive charges, and in that case the negative charges would be more densely packed by a factor of $$γ$$ when the current was moving – but also when it wasn't. They would be denser by a factor of $$γ$$ not because of length contraction, but because you shoved $$γ$$ times as many negative charges as positive charges into the wire before you closed the circuit. If you look at only a straight subsection of that wire, then your three bulleted cases are basically correct. If the positive test particle moves in the direction of the negative current (that is, opposite the current vector) then there is a repulsive magnetic force which matches and then exceeds the attractive electric force. Note that "half the velocity" is not $$v/2$$ but rather the solution to $$2u/(1+u^2)=v$$ (or half the rapidity).

Repeating my comment... My biggest problem with that segment between 1:20 and 2:30 is that the presenter starts with an overall neutral wire... but the instant the negative charges start moving (at 1:28), he has actually changed the physical situation entirely, and switched to a wire that is neutral only when negative charges are moving, but positive overall if the current stops. The wires before 1:28 and after 1:28 are completely different wires.

To describe the physics of your 3 scenarios:

1. Test particle is stationary, sees stationary positive charges and moving + contracted negative charges. The negative charge density is greater, so the test charge feels a force toward the wire.

(Assuming the frame of the test charge.) You are right that the contracted -ve charges cause the +ve test charge to be pulled toward the wire... due to ELECTRIC force. Keep in mind the moving -ve charges do create a magnetic field, but since the test charge is stationary, the magnetic field doesn't do anything to it.

1. Test particle moves in the direction of the negative current, but with half the velocity of the current. This setup results in the test particle seeing both the negative and positive charges in the wire appearing to move away from the test particle with the same velocity but opposite directions. So both would be equally contracted in distance, resulting in equal charge densities, leading to no force on the test particle.

(Assuming the frame where the charge is moving at half the speed of the -ve charges.) Your interpretation is not quite correct. You are right that the positive and negative charge densities are both contracted equally, so there is no ELECTRIC force on the test charge. However, the moving +ve charges and moving -ve charges do both create MAGNETIC fields. And most importantly, these magnetic fields are not opposites that cancel... they are in the same direction and actually add together (this is because the opposite charges are moving in opposite directions... so their magnetic fields are aligned). So the test particle feels a MAGNETIC force pulling it towards the wire.

1. Test particle moves along at the same velocity as the negative charge, so this charge is not contracted, but now the positive charges in the wire are contracted and as much so as the negative charges were in the first scenario. So now the positive charge density is greater, and the particle feels a force away from the wire.

(Assuming the frame of the test charge.) The +ve charges are indeed length contracted and denser than the -ve charges, so the test charge does does get pushed away by an ELECTRIC force.

To understand the relativistic view of this problem, it's better to do a thought experiment with a thought experiment type set up. That is: keep it simple. There is no need to worry about how the current gets started, or collisions and all that real stuff.

So, imagine a uniformly spaced lattice of positive charges, with spacing $$a=1$$. Each positive charge is balanced by a negative charge moving to the right at same speed, say with $$\gamma = 2$$.

In the figure below, I have show 5 ion/electron pairs, in blue/red for the unprimed/primed frame:

Of course, the ions are stationary, so their world-lines are vertical. The electrons move to the right. An $$x-$$slice at $$t=0$$ shows that they are neutral, each with a spacing of $$a=1$$, that is, the charge densities are:

$$\rho_{\pm} = \frac {\pm e} a$$

so that:

$$\rho_+ +\rho_-=0$$

The symbols are placed at uniform temporal spacing in the unprimed frame (except for the magenta, showing the length $$a=1$$ in the $$x'$$ frame).

Now boost each point to the primed frame:

In this frame, the electron word-lines are vertical, with a dilated spacing of $$\gamma a = 2$$. The ion world-lines move to the left, and the spacing is Lorentz contracted to $$a/\gamma = \frac 1 2$$. The charge densities are:

$$\rho'_{\pm} = \gamma^{\pm 1}\frac {\pm e} a$$

for a net charge density of:

$$\rho'_+ +\rho'_-= (\gamma - \frac 1 {\gamma} )\frac e a$$

The biggest point of confusion is generally, "why don't the moving electrons Lorentz contract in the wire frame?". Why should they? They are independent objects moving to the right: there is nothing to contract. Meanwhile, in the primed frame, the lattice of ions does contract. The ions are bound together in the wire.

Of course, since the electrons don't contract in the lab frame, the must be dilated in their own rest frame. For the electrons to accelerate synchrnously in the lab frame, while maintaining uniform spacing, the must spread out in their own rest frame, which is similar to Bell's Spaceship Paradox. If a thin thread joined each electron to its neighbors, choose threads would break when the current is turned on.