Time intervals measured by stationary and moving observers

[Hak]

[Moderator's note: Thread moved to relativity forum as it is not a specific homework problem but a general question about SR.]

Homework Statement: While studying relativity, a question arose for me about time intervals measured by stationary and moving observers. In particular, one of the first steps after clarifying the two postulates is to prove that $$dS^2 = ds^2$$, i.e. that the square of the distance between two events is invariant. Now let us consider a reference system $$K$$ moving with constant velocity with respect to a second reference system $$k$$. By virtue of one of the two postulates, the observer in solidarity with the $$K$$ reference system cannot distinguish whether he or the $$k$$ reference system is moving. He can therefore assume that he is at rest, and therefore claim that $$c^2 dT^2 = c^2 dt^2 - dx^2$$, where dx is the distance that the $$K$$ reference system claims is travelled by the $$k$$ system. From this equation we arrive at saying that $$dT< dt$$.
If, however, $$K$$ were considered to be in motion, the former would become $$c^2 dT^2 - dX^2= c^2 dt^2$$, from which it follows that $$dT> dt$$. Could you explain to me why the two results are conflicting?
Relevant Equations: /

The only thing I can think of is that really $$K$$ cannot say anything about the interval measured in another reference system, unless there is an asymmetry (like the twin on the spaceship in the relevant paradox, in which he is forced to return to his brother to check that his watch is lagging behind the clock on the ground and in making this return is forced to decelerate until he reverses his speed, realising that he is the one in the system in motion). Thank you.

 

[kuruman]

Are familiar with relativity of simultaneity?

 

[TSny]

Could you explain to me why the two results are conflicting?

The two results are not in conflict since the two results are not referring to the same events.

In the first case you are considering two events ($a$ and $b$, say) that occur at the same spatial point but at different times in frame K. That is, $dX_{(a,b)} = 0$ and $dT_{(a,b)} \neq 0$. These two events occur at different spatial points and different times in frame k: $dx_{(a,b)} \neq 0$ and $dt_{(a,b)} \neq 0$. The subscripts $(a, b)$ emphasize reference to the two events $a$ and $b$.

In the second case you are considering two events $c$ and $d$ for which $dx_{(c, d)} = 0$, $dt_{(c, d)} \neq 0$, $dX_{(c, d)} \neq 0$, and $dT_{(c, d)} \neq 0$.

It is not hard to see that the two events $a$ and $b$ cannot be the same events as $c$ and $d$. So there is no conflict in claiming that $dT_{(a,b)} < dt_{(a,b)}$ while $dt_{(c, d)} < dT_{(c, d)}$.

 

[Hak]

Are familiar with relativity of simultaneity?

Unfortunately not, or at least not that well. I have heard some information, but I do not know the principles in detail. Could you explain in more detail what it is and how it applies to my previous comment?

 

[Hak]

The two results are not in conflict since the two results are not referring to the same events.

In the first case you are considering two events ($a$ and $b$, say) that occur at the same spatial point but at different times in frame K. That is, $dX_{(a,b)} = 0$ and $dT_{(a,b)} \neq 0$. These two events occur at different spatial points and different times in frame k: $dx_{(a,b)} \neq 0$ and $dt_{(a,b)} \neq 0$. The subscripts $(a, b)$ emphasize reference to the two events $a$ and $b$.

In the second case you are considering two events $c$ and $d$ for which $dx_{(c, d)} = 0$, $dt_{(c, d)} \neq 0$, $dX_{(c, d)} \neq 0$, and $dT_{(c, d)} \neq 0$.

It is not hard to see that the two events $a$ and $b$ cannot be the same events as $c$ and $d$. So there is no conflict in claiming that $dT_{(a,b)} < dt_{(a,b)}$ while $dt_{(c, d)} < dT_{(c, d)}$.

Thank you so much.

 

[PeroK]

Thank you so much.

To prove that the spacetime interval is invariant across all inertial reference frames, you must first derive the Lorentz Transformation. This is the transformation of spacetime coordinates between two inertial reference frames. Note that the Lorentz Transformation encodes the three basic concepts of time dilation, length contraction and relativity of simultaneity.

See, for example:

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html

 

[Hak]

To prove that the spacetime interval is invariant across all inertial reference frames, you must first derive the Lorentz Transformation. This is the transformation of spacetime coordinates between two inertial reference frames. Note that the Lorentz Transformation encodes the three basic concepts of time dilation, length contraction and relativity of simultaneity.

See, for example:

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html

Thank you so much. I found out that document. Is it helpful about what you recommend and concerning what I would like clarified?

 

[Hak]

While I am at it I use that section to ask another question with respect to the theory. This time the doubt arises with respect to relativistic action, a not exactly minor aspect since it leads to the conclusion that bodies follow geodesics of space-time. I quote it:
$$S = -\alpha \int_{a}^{b} ds$$
Where $$\alpha$$ represents a positive constant to be determined by comparison with the classical Lagrangian of a free particle. It does not make explicit the regions of the minus sign, but it is essential for the action to have a minimum. The question is this: how is it that the very integral is minimized with respect to $$ds$$? One clever justification I have read is that the action must be invariant with respect to Lorentz transformations. The problem is that I don't understand why exactly $$ds$$ was chosen as the scalar.
I also try in this to give a possible explanation, but one that I do not know in the slightest whether it is correct: By positing $$ds$$ as the scalar and expressing the latter in function $$dt$$ and after finding the value of $$\alpha$$ by comparison with the classical lagrangian, one obtains the relativistic lagrangian. Once the latter is derived as a function of the velocity, one obtains as is known the momentum of the particle. The latter value could have been derived by other means without pulling in the Lagrangian and, therefore, retracing the path backward considering the particle's momentum known yields the action written at the beginning. Does what I am saying make sense or am I wrong and is there a deeper reason? Thanks again.

 

[PeroK]

The question is this: how is it that the very integral is minimized with respect to $$ds$$? One clever justification I have read is that the action must be invariant with respect to Lorentz transformations.

$ds$ is the differential spacetime distance, so $ds$ is invariant under any Lorentz Transformation. Or, more directly$$\int_a^b ds = \int_a^b ds'$$Where $a, b$ represent two spacetime events. This is because both integrals represent the spacetime distance between $a$ and $b$.

 

[Hak]

$ds$ is the differential spacetime distance, so $ds$ is invariant under any Lorentz Transformation. Or, more directly$$\int_a^b ds = \int_a^b ds'$$Where $a, b$ represent two spacetime events. This is because both integrals represent the spacetime distance between $a$ and $b$.

Thank you very much, but why $ds$ was chosen as a scalar? Could you please help me clarifying some doubts about the part of my message that you did not mention?

 

[PeroK]

Thank you very much, but why $ds$ was chosen as a scalar? Could you please help me clarifying some doubts about the part of my message that you did not mention?

There's a derivation/justification here:

https://kevinshuang.com/2017/11/25/lagrangian-for-a-relativistic-free-particle/

 

[Hak]

There's a derivation/justification here:

https://kevinshuang.com/2017/11/25/lagrangian-for-a-relativistic-free-particle/

Forgive me, I cannot understand how in this derivation/justification, it is explained why $ds$ is a scalar. Thank you for your patience.

 

[PeroK]

Forgive me, I cannot understand how in this derivation/justification, it is explained why $ds$ is a scalar. Thank you for your patience.

The definition of a scalar is that it is invariant under coordinate transformations. The spacetime interval, therefore, is a scalar: in finite form $\Delta s$ or differential form $ds$.

 

[Hak]

Thank you indeed!

 

[Hak]

There's a derivation/justification here:

https://kevinshuang.com/2017/11/25/lagrangian-for-a-relativistic-free-particle/

I have viewed this justification; it seems quite solid and convincing. It is all clear to me except for the following development, which is equation (3) of this explanation to which you referred me.

The new Lagrangian $L(v'^2)$ is given by:

$$L(v'^2)=L(v^2)+\frac{\partial L}{\partial (v^2)}2vV\left(1-\frac{v^2}{c^2}\right)$$.

It's probably trivial, but I can't figure it out even for classical Lagrangian. Could you please explain this passage to me?

 

[Hak]

I have viewed this justification; it seems quite solid and convincing. It is all clear to me except for the following development, which is equation (3) of this explanation to which you referred me.

The new Lagrangian $L(v'^2)$ is given by:

$$L(v'^2)=L(v^2)+\frac{\partial L}{\partial (v^2)}2vV\left(1-\frac{v^2}{c^2}\right)$$.

It's probably trivial, but I can't figure it out even for classical Lagrangian. Could you please explain this passage to me?

Is this an assumption, or is there a specific motivation that leads to such a relationship?

 

[PeroK]

I have viewed this justification; it seems quite solid and convincing. It is all clear to me except for the following development, which is equation (3) of this explanation to which you referred me.

The new Lagrangian $L(v'^2)$ is given by:

$$L(v'^2)=L(v^2)+\frac{\partial L}{\partial (v^2)}2vV\left(1-\frac{v^2}{c^2}\right)$$.

It's probably trivial, but I can't figure it out even for classical Lagrangian. Could you please explain this passage to me?

The idea is that for small $V$ we have:
$$(v')^2 \approx v^2 + 2vV(1 - \frac{v^2}{c^2})$$And, using the Taylor expansion for $L$ at the point $(v')^2$:
$$L((v')^2) = L(v^2) + \frac{\partial L(v^2)}{\partial (v^2)}\big ((v'^2) - v^2) \big)$$And the result follows.

 

[Hak]

The idea is that for small $V$ we have:
$$(v')^2 \approx v^2 + 2vV(1 - \frac{v^2}{c^2})$$And, using the Taylor expansion for $L$ at the point $(v')^2$:
$$L((v')^2) = L(v^2) + \frac{\partial L(v^2)}{\partial (v^2)}\big ((v'^2) - v^2) \big)$$And the result follows.

Ok, thank you. Where, however, does the latest formula come from? What does the Taylor expansion need to be applied to? Could you show me some calculations to that effect? Thank you.

 

[PeroK]

Ok, thank you. Where, however, does the latest formula come from? What does the Taylor expansion need to be applied to? Could you show me some calculations to that effect? Thank you.

That's an exercise for you. $v'$ is a function of $v$, with $V$ as a parameter, as given by the velocity transformation given in the text.

 

[Hak]

The idea is that for small $V$ we have:
$$(v')^2 \approx v^2 + 2vV(1 - \frac{v^2}{c^2})$$And, using the Taylor expansion for $L$ at the point $(v')^2$:
$$L((v')^2) = L(v^2) + \frac{\partial L(v^2)}{\partial (v^2)}\big ((v'^2) - v^2) \big)$$And the result follows.

Recall that the Taylor series of a function $f(x)$ about a point $x=a$ is given by: $$f(x) = f(a) + f'(a) \ (x-a) + \frac{f''(a)}{2!} (x-a)^2 + ...$$.
If we let $f(x)=L(x)$, $x=(v')^2$, and $a=v^2$, and if we assume that $V$ is small enough, then we can neglect the higher order terms in the series and obtain: $$L((v')^2) = L(v^2) + L'(v^2)\big ((v'^2) - v^2) \big)$$. Calculations will continue from here, with: $L'(v^2) = \frac{\partial L(v^2)}{\partial (v^2)}$.
Is it correct?

 

[Hak]

The idea is that for small $V$ we have:
$$(v')^2 \approx v^2 + 2vV(1 - \frac{v^2}{c^2})$$And, using the Taylor expansion for $L$ at the point $(v')^2$:
$$L((v')^2) = L(v^2) + \frac{\partial L(v^2)}{\partial (v^2)}\big ((v'^2) - v^2) \big)$$And the result follows.

Recall that the Taylor series of a function $f(x)$ about a point $x=a$ is given by: $$f(x) = f(a) + f'(a) \ (x-a) + \frac{f''(a)}{2!} (x-a)^2 + ...$$.
If we let $f(x)=L(x)$, $x=(v')^2$, and $a=v^2$, and if we assume that $V$ is small enough, then we can neglect the higher order terms in the series and obtain: $$L((v')^2) = L(v^2) + L'(v^2)\big ((v'^2) - v^2) \big)$$. Calculations will continue from here, with: $L'(v^2) = \frac{\partial L(v^2)}{\partial (v^2)}$.
Is it correct?

@PeroK Isn't there an inaccuracy here? Isn't it more appropriate to define it so that $x = v^2$, $a = v'^2$, so that the development of $L(v^2)$ calculated at the point $L(v'^2)$ becomes:

$$L(v^2) = L(v'^2) + L'(v'^2) (v^2 - v'^2)$$?And at this point, with a simple algebraic step:

$$L(v'^2) = L(v^2) + L'(v'^2) (v'^2 - v^2)$$.

Which of the two definitions of $x$ and $a$ are correct? The ones I just stated or the ones previously explicated? Thank you.

 

[PeroK]

Which of the two definitions of $x$ and $a$ are correct? The ones I just stated or the ones previously explicated? Thank you.

The previous one was right.

 

[Hak]

The previous one was right.

Could you explain why in detail? I don't think I understood, I thought the second ones were right...

 

[PeroK]

Could you explain why in detail. I don't think I understood, I thought the second ones were right...

The derivatives in a Taylor series are evaluated at the reference point (in this case at $v^2$), not at the variable point (in this case $v'^2$).

 

[Hak]

The derivatives in a Taylor series are evaluated at the reference point (in this case at $v^2$), not at the variable point (in this case $v'^2$).

So this means that $x = v'^2$ and $a = v^2$, right? Could you direct me to an article that explains this question about Taylor series well, so that I can understand it better? Thank you very much.

 

[PeroK]

Google is your friend.

 

[Hak]

Google is your friend.

I tried searching, but I couldn't find anything interesting. Maybe I didn't search well within it....

 

[Hak]

I tried searching, but I couldn't find anything interesting. Maybe I didn't search well within it....

I would like to tap into some reliable sources that explain this particular aspect of Taylor series in detail. I haven't found any on the net.

 

[Hak]

Sorry if I did not understand what @PeroK said. I can't understand why $v^2$ is the reference point (thus the evaluation point), while $v'^2$ is the variable one. I would like to know on what basis it is defined or decreed, if I am not asking too much. From what, then, comes the decision to calculate the derivative in Taylor development at the reference point and not the variable one. Why? I would be grateful if you could let me know. Thank you very much.

 

[PeterDonis]

From what, then, comes the decision to calculate the derivative in Taylor development at the reference point and not the variable one.

The definitions of "reference point" and "variable point" are that the reference point is where you calculate the derivative and the variable point is not.

Note that you could switch frames and do the calculation in the primed frame; then $v'$ would be the reference point and $v$ would be the variable point. Either choice is fine, but once you've made the choice you have to apply it consistently.

 

[Hak]

The definitions of "reference point" and "variable point" are that the reference point is where you calculate the derivative and the variable point is not.

Note that you could switch frames and do the calculation in the primed frame; then $v'$ would be the reference point and $v$ would be the variable point. Either choice is fine, but once you've made the choice you have to apply it consistently.

Thank you very much for your reply. What do you mean by "you have to apply it consistently"? Thank you again.

 

[PeterDonis]

What do you mean by "you have to apply it consistently"?

I mean that once you have picked a reference frame, that defines for you which point is the "reference point" and which point is the "variable point". You don't have any choice then about which is which.

 

[Hak]

I mean that once you have picked a reference frame, that defines for you which point is the "reference point" and which point is the "variable point". You don't have any choice then about which is which.

Thank you very much. Sure, I agree. I am still, however, confused. When I had asked which definition of $x$ and $a$ was correct, @PeroK replied "$x = v'^2$ and $a = v^2$", not that the interchanged ones (i.e. $x = v^2$ and $a = v'^2$) were fine too. In the light of what you told me, why are my calculations incorrect in post #21? Have I not been consistent with my choice of reference? Try looking at post #21 and, if you can and want to, let me know. Thank you very much again.

 

[PeterDonis]

When I had asked which definition of $x$ and $a$ was correct, @PeroK replied "$x = v'^2$ and $a = v^2$"

That's because he knew you were using the unprimed reference frame.

not that the interchanged ones (i.e. $x = v^2$ and $a = v'^2$) were fine too.

That's because they're not fine if you have already chosen the unprimed reference frame.

In the light of what you told me, why are my calculations incorrect in post #21? Have I not been consistent with my choice of reference?

Obviously not, because in post #21 you switched reference frames (you didn't realize that's what you were doing when you said $x = v^2$, $a = v'^2$, but it was).

 

[Hak]

That's because he knew you were using the unprimed reference frame.That's because they're not fine if you have already chosen the unprimed reference frame.Obviously not, because in post #21 you switched reference frames (you didn't realize that's what you were doing when you said $x = v^2$, $a = v'^2$, but it was).

Thank you. Yes, I wrote $L' (v'^2)$ instead of $L'(v^2)$, right?