Can the number 1 be squared infinitely many times?

[rgtr]

Homework Statement

When calculating relativity of simultaneity I am confused by one thing

Relevant Equations

$\left( \frac {LV} {C^2} \right)$

When calculating relativity of simultaneity I am confused by one thing. How is the math = to the answer below?

$\left( \frac {LC+LV } {2C} \right) - \left( \frac {LC -LV} {2C} \right) = \left( \frac {LV} {C^2} \right)$

The part that throwing me off is $C^2$ . I assume mathematically since c = 1 I can mathematically add a squared of any number to +1?
I realize this a stupid question but just want to confirm.

 

[Orodruin]

You have not specified what it is you are trying to compute. In general, if you are using c=1 it should not appear at all. If you want to go from c=1 to units where this is not the case you can do so by dimensional analysis.

 

[rgtr]

I mean I can square 1 as many times as I want. so 1 can be $1^2$ or $1 ^{infinite number }$

 

[Orodruin]

I mean I can square 1 as many times as I want. so 1 can be $1^2$ or $1 ^{infinite number }$

But only one of the options will be dimensionally consistent when you reintroduce $c\neq 1$.

 

[rgtr]

Okay thanks but it is possible in relativity of simultaneity because I get $\frac {LV} {C}$ ?

 

[Orodruin]

Okay thanks but it is possible in relativity of simultaneity because I get $\frac {LV} {C}$ ?

You have still not stated what this is supposed to describe. Without that it is impossible to answer.

 

[Ibix]

$\left( \frac {LC+LV } {2C} \right) - \left( \frac {LC -LV} {2C} \right) = \left( \frac {LV} {C^2} \right)$

Are you asking, in short, "is it legitimate to simplify the LHS to $\frac{LV}{c}$ and then introduce an extra factor of $c$ in the denominator to make it equal the RHS, given $c=1$"? If so, no it isn't. You either don't put $c$ in at all and only reintroduce it when necessary (the $c=1$ convention), or you keep track of it like any other algebraic symbol.

Something has gone wrong in your algebra somewhere. You haven't shown us it, so we don't know what.

 

[rgtr]

Here is the book I am using.

https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf (page 15)

I am confused how it starts at $\frac {LV} {C}$ to $\frac {LV} {C^2}$. How does that work?

Let’s now examine what happens during the process in the train frame. Compared
with the forward-moving light, Fig. 1.12 tells us that the backward-moving light must
travel an extra distance of L(c + v)/2c − L(c − v)/2c = Lv/c. The light travels at speed c
(as always), so the extra time is Lv/c2. The rear clock therefore reads Lv/c2 more when
it is hit by the backward photon,
Also why is the result above different then the video below?

 

[bob012345]

Homework Statement:: When calculating relativity of simultaneity I am confused by one thing
Relevant Equations:: (LVC2)

When calculating relativity of simultaneity I am confused by one thing. How is the math = to the answer below?

(LC+LV2C)−(LC−LV2C)=(LVC2)

The part that throwing me off is C2 . I assume mathematically since c = 1 I can mathematically add a squared of any number to +1?
I realize this a stupid question but just want to confirm.

I am missing something in this algebra. How did you get this formula?
$$\left( \frac {LC+LV } {2C} \right) - \left( \frac {LC -LV} {2C} \right) = \left( \frac {LV} {C^2} \right)$$and shouldn't it be?
$$\left( \frac {LC+LV } {2C} \right) - \left( \frac {LC -LV} {2C} \right) = \left( \frac {LV} {C} \right)$$

I think keeping track of relativity is hard enough for students that units of c=1 should be left to the experts because dimensions can too easily get messed up.

 

[nasu]

@rgtr There is no such formula (as you wrote it in the OP) in the text given as reference . The result of simplifying the left hand side is with "c" and not c². Only later, this result (which has dimensions of distance) is divided by c in order to get a time difference. It is quite clear in the text on page 15. It would make more sense to ask directly the question you have rather than make up some scenario with infinite power. Or read again if you have some misunderstanding.

 

[Orodruin]

the backward-moving light must
travel an extra distance of L(c + v)/2c − L(c − v)/2c = Lv/c. The light travels at speed c
(as always), so the extra time is Lv/c2.

If it has to travel a distance x more and travels at speed c, then clearly that takes x/c more time. There is no c=1 issue going on here at all.

 

[Orodruin]

I think keeping track of relativity is hard enough for students that units of c=1 should be left to the experts because dimensions can too easily get messed up.

I have the opposite standpoint. Relativity is hard enough without having factors of c popping up everywhere to obscure the actual structure of the theory.

 

[bob012345]

I have the opposite standpoint. Relativity is hard enough without having factors of c popping up everywhere to obscure the actual structure of the theory.

I view c as the whole point of the theory. What is the actual structure as you see it?

 

[Orodruin]

I view c as the whole point of the theory. What is the actual structure as you see it?

Pseudo Riemannian geometry.

 

[Ibix]

I view c as the whole point of the theory. What is the actual structure as you see it?

Seeing relativity as a theory of spacetime geometry, $c$ is just a conversion factor between units of time and units of distance. Thus setting $c\neq 1$ makes as much sense as using fathoms for depth and nautical miles for horizontal distance. It works when there is no cross-talk between the two, as in Newtonian physics (or ship navigation in the analogy), but just obscures important stuff when you can mix them, as in relativity (or if you want to ask how far a submarine at 1000 fathoms has to travel if it wants to surface 3 nautical miles away).

Sailors might use fathoms and nautical miles anyway because they're more practical in every day use and depth charts are probably labelled in fathoms for legacy reasons. But fundamental science cares less about everyday practicality and more about showing the symmetries we've found useful.

 

[Orodruin]

Thus setting c≠1 makes as much sense as using fathoms for depth and nautical miles for horizontal move

I was just starting to write this as the page reloaded …

 

[Orodruin]

I mean, it makes sense in a practical way but it obscures the fact that the dimensions are fundamentally related.

 

[Ibix]

I was just starting to write this as the page reloaded …

It's not an original analogy on here...

I mean, it makes sense in a practical way but it obscures the fact that the dimensions are fundamentally related.

Absolutely. I added a bit more to my last post (I posted early by accident) trying to make this point.

 

[Ibix]

Here is the book I am using.

https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf (page 15)

I am confused how it starts at $\frac {LV} {C}$ to $\frac {LV} {C^2}$. How does that work?

As @nasu points out, $LV/c$ is the "extra" distance and therefore $LV/c^2$ is the flight time difference for something traveling at $c$.

Let’s now examine what happens during the process in the train frame. Compared
with the forward-moving light, Fig. 1.12 tells us that the backward-moving light must
travel an extra distance of L(c + v)/2c − L(c − v)/2c = Lv/c. The light travels at speed c
(as always), so the extra time is Lv/c2. The rear clock therefore reads Lv/c2 more when
it is hit by the backward photon,

You didn't indicate that this paragraph is copy-pasted from the book. In many universities that would get you a plagiarism warning. I don't think we are quite so strict at PF, but it is helpful to indicate which bits of text are quotations you wish to discuss and which are your discussion.

 

[Orodruin]

A few years back I had a new TA in my special relativity course. At the first meeting we had a lengthy discussion about what constants were ”natural” to put equal to one. Coming from condensed matter theory he was basically against everything except $k_B=1$. By the end of the course he was very happily endorsing both $c=1$ and $\hbar=1$.

 

[bob012345]

Pseudo Riemannian geometry.

In my mind I was just limiting the discussion to SR and practical problems for beginning students which I interpreted as being more in line with the OP. That's what I meant by suggesting using c=1 is fine for pros or those advanced enough to be concerned with GR.

 

[Orodruin]

That's what I meant by suggesting using c=1 is fine for pros or those advanced enough to be concerned with GR.

Minkowski space is also pseudo Riemannian. It is to a general pseudo Riemannian manifold what Euclidean space is to a Riemannian one. There is a lot of useful geometry going into it. I can understand not letting c = 1 in "modern physics" style courses or very introductory overviews, but as soon as you become only a little more advanced then c = 1 is clearly preferable. For example, I use c = 1 in my SR course at master level, it is definitely not only for "pros".

 

[rgtr]

As @nasu points out, $LV/c$ is the "extra" distance and therefore $LV/c^2$ is the flight time difference for something traveling at $c$.

You didn't indicate that this paragraph is copy-pasted from the book. In many universities that would get you a plagiarism warning. I don't think we are quite so strict at PF, but it is helpful to indicate which bits of text are quotations you wish to discuss and which are your discussion.

I am sorry I didn't mean to pass on the paragraph as mine. I linked the book I just forgot to put the text in
quotes. I thought it was implied because I thought someone would click on the link first before preceding.

Also I understand c = 1 but can be different. It is just in certain instances c = 1, that is all I meant.
Also what is flight time difference ?

 

[Ibix]

I am sorry I didn't mean to pass on the paragraph as mine. I linked the book I just forgot to put the text in
quotes.

I realized after I read the link. I was just pointing out that that particular piece of carelessness can get you into surprisingly deep trouble in some contexts.

Also what is flight time difference ?

In all frames except one the forward and backward going light pulses travel different distances. The time they take is the flight time of the light. The difference between the times is the flight time difference.

 

[rgtr]

So essentially what you are saying is
$d = \frac {Lv} {c}$$t = \frac {d} {v}$$t = \frac {Lv} {\frac {c} {c} } = \frac {lv} {c^2}$

I should have got that sooner. Duh. I am not the greatest reader so I assumed the distance was time. Sorry.

 

[haruspex]

So essentially what you are saying is
$d = \frac {Lv} {c}$$t = \frac {d} {v}$$t = \frac {Lv} {\frac {c} {c} } = \frac {lv} {c^2}$

I should have got that sooner. Duh. I am not the greatest reader so I assumed the distance was time. Sorry.

I can't make sense of your algebra. Did you mean
$d = \frac {Lv} {c}$
$t = \frac {d} {c}$
$t = \frac {(\frac {Lv} {c})} {c} = \frac {Lv} {c^2}$
?

 

[rgtr]

I can't make sense of your algebra. Did you mean
$d = \frac {Lv} {c}$
$t = \frac {d} {c}$
$t = \frac {(\frac {Lv} {c})} {c} = \frac {Lv} {c^2}$
?

Yep. How is mine much different? Sorry for dragging this conversation .

 

[haruspex]

Yep. How is mine much different? Sorry for dragging this conversation .

You wrote $t = \frac {d} {\textbf v}$ Instead of $t = \frac {d} {\textbf c}$
and $\frac {Lv} {\frac {c} {c} }$
which would reduce to $Lv$

 

[rgtr]

I still don't see how the 3rd term is different between you and me in post 25 and 26? Again I am sorry for the stupid question?

 

[haruspex]

I still don't see how the 3rd term is different between you and me in post 25 and 26? Again I am sorry for the stupid question?

Your nesting of the fractions makes it read like $\frac {Lv} {(\frac {c} {c}) }$