Deriving Lorentz Transformations: Hyperbolic Functions

[Samama Fahim]

While deriving Lorentz transformation equations, my professor assumes the following:

As $\beta \rightarrow 1,$

$$-c^2t^2 + x^2 = k$$

approaches 0. That is, $-c^2t^2 + x^2 = 0.$ But the equation of the hyperbola is preserved in all inertial frames of reference. Why would $-c^2t^2 + x^2$ become zero instead of remaining $k \neq 0$ if $\beta$ approaches 1? Does that make sense in any other way? He relates that with the following equation:

$$\gamma^2 -\beta^2\gamma^2 = 1$$

 

[PeroK]

While deriving Lorentz transformation equations, my professor assumes the following:

As $\beta \rightarrow 1,$

$$-c^2t^2 + x^2 = k$$

approaches 0. That is, $-c^2t^2 + x^2 = 0.$ But the equation of the hyperbola is preserved in all inertial frames of reference. Why would $-c^2t^2 + x^2$ become zero instead of remaining $k \neq 0$ if $\beta$ approaches 1? Does that make sense in any other way? He relates that with the following equation:

$$\gamma^2 -\beta^2\gamma^2 = 1$$

I guess $t$ and $x$ are the trajectory of a particle moving at speed $v$, starting at the origin. In which case we have $x = vt$ and:
$$-c^2 t^2 + x^2 = -c^2t^2 + v^2t^2 = (\beta^2 - 1)c^2t^2$$

 

[Samama Fahim]

I guess $t$ and $x$ are the trajectory of a particle moving at speed $v$, starting at the origin. In which case we have $x = vt$ and:
$$-c^2 t^2 + x^2 = -c^2t^2 + v^2t^2 = (\beta^2 - 1)c^2t^2$$

That makes sense. But it still does not make $-c^2 t^2 + x^2$ a constant.

 

[PeroK]

That makes sense. But it still does not make $-c^2 t^2 + x^2$ a constant.

The expression $-c^2t^2 + x^2$ is generally frame invariant. That's different from being a constant. Suppose we take any point on the hyperbola $-c^2t^2 + x^2 = k$, where $(t, x)$ are the coordinates in some inertial reference frame and we transform those coordinates to a different inertial frame $(t', x')$. Then we find that $-c^2t'^2 + x'^2 = k$.

In some sense, therefore, that hyperbola is frame invariant.

Without knowing the approach your professor is taking that's about as much as I can say.

 

[Samama Fahim]

But the equation $-c^2t^2+x^2 = k$ cannot describe a particle having a constant speed $v.$

 

[PeroK]

But the equation $-c^2t^2+x^2 = k$ cannot describe a particle having a constant speed $v.$

That's right. Unless the particle is massless, hence $v = c$ and $k = 0$. In which case you have a pair of straight lines, which is a degenerate hyperbola.

 

[Orodruin]

While deriving Lorentz transformation equations, my professor assumes the following:

As $\beta \rightarrow 1,$

$$-c^2t^2 + x^2 = k$$

approaches 0. That is, $-c^2t^2 + x^2 = 0.$

No it doesn't. The hyperbola is frame invariant. Perhaps it would be better if you could state word by word what your professor says or, even better, provide a link to the lecture notes.

 

[PeterDonis]

While deriving Lorentz transformation equations, my professor assumes the following:

As $\beta \rightarrow 1,$

$$-c^2t^2 + x^2 = k$$

approaches 0. That is, $-c^2t^2 + x^2 = 0.$

This will depend on how the limit is taken. Your professor appears to be implicitly assuming that the limit is being taken along a straight line of constant $t$, moving from $x = 0$ (where $\beta = 0$) towards the limit point $x = 1$ (where $\beta = 1$). While this is a perfectly valid procedure mathematically, it doesn't correspond to anything useful physically. Nor do I see how this limit would be relevant for deriving the Lorentz transformation equations.

Physically, if you imagine an object being accelerated from $\beta = 0$ towards $\beta = 1$ with constant proper acceleration, that object's worldline in spacetime will be a hyperbola that satisfies the equation $-c^2t^2 + x^2 = k$ for a constant $k$, as you say. Since the Lorentz transformations will leave this hyperbola invariant in the sense that @PeroK described in post #4, one could use this, I suppose, in a derivation of the Lorentz transformation equations, although I don't think it's the simplest way to do that.

 

[Orodruin]

one could use this, I suppose, in a derivation of the Lorentz transformation equations, although I don't think it's the simplest way to do that.

Considering that it is just identifying the hyperbolic one two times and then the hyperbolic sine of a sum ... I’d call it pretty simple actually.

 

[Samama Fahim]

What do you mean by 'identifying the hyperbolic one two times'?

 

[PeroK]

What do you mean by 'identifying the hyperbolic one two times'?

I think we need some explanation from you of what your are trying to do here. Vague quotes from your professor just lead to confusion.

 

[Samama Fahim]

I think we need some explanation from you of what your are trying to do here. Vague quotes from your professor just lead to confusion.

Just trying to make sense of what the professor did. When $-c^2t^2+x^2=0,$ $v = \frac{x}{t} = c.$ From this he shows that $\beta = \frac{v}{c}.$

 

[PeroK]

Just trying to make sense of what the professor did. When $-c^2t^2+x^2=0,$ $v = \frac{x}{t} = c.$ From this he shows that $\beta = \frac{v}{c}.$

I suspect there's more to it that you are not telling us.

 

[PeterDonis]

When $-c^2t^2+x^2=0,$ $v = \frac{x}{t} = c.$ From this he shows that $\beta = \frac{v}{c}.$

How? This is not enough detail to know what your professor did or to be able to evaluate it.

We need a link to the actual lecture notes, or something equivalent.

 

[PeterDonis]

When $-c^2t^2+x^2=0,$ $v = \frac{x}{t} = c.$ From this he shows that $\beta = \frac{v}{c}.$

As you write it, this doesn't even make sense. $\beta = v / c$ is a definition. It's not something you "show". In the OP to this thread you said the professor was deriving the Lorentz transformation equations.

 

[Samama Fahim]

We started with the following assumptions:
$$\gamma = \cosh(\phi)$$
$$\beta = \tanh(\phi)$$.

 

[PeterDonis]

We started with the following assumptions:
$$\gamma = \cosh(\phi)$$
$$\beta = \tanh(\phi)$$.

We are not asking for what the professor started with. Either you need to post a link to the actual lecture notes, or whatever other source the professor is using, that contains the full derivation you are referring to, or you need to post the actual full derivation here. Expecting us to comment on every single step one at a time is not reasonable. Either do one of the above or this thread will be closed, as we cannot help you without that information.

 

[Samama Fahim]

OK I am posting the full derivation here.

 

[Orodruin]

What do you mean by 'identifying the hyperbolic one two times'?

If $A^2 - B^2 = 1$ then this can be parametrised as $A = \cosh^2\theta$ and $B = \sinh^2\theta$ for some $\theta$.

 

[Samama Fahim]

We want to preserve the dot product (actually the hyperbola):

$$-c^2t^2+x^2=k.$$

This suggests that the transformation that would preserve the hyperbola can be represented by

$$\cosh^(\phi) - \sinh^2(\phi) = 1.$$

Let $\gamma = \cosh(\phi)$ and $\beta = \tanh(\phi),$ then

$\gamma^2 - \beta^2 \gamma^2 = 1.$

'Now what is the nature of $\beta$?' asks the professor.

Well, it should be dimensionless since you cannot add a physical quantity ($\beta^2$) to a pure number (1):

$\gamma = \frac{1}{\sqrt{1-\beta^2}}.$

He goes on to write

Since

$$-c^2t^2+x^2=k$$
$$\gamma^2-\beta^2 \gamma^2 = 1.$$

Then if $\beta$ approaches 1, $-c^2t^2+x^2$ approaches 0. (I have no idea how he relates these two equations. It could be that if $\beta$ is the ratio $v/c$, then $\beta$ approaching 1 means that the particle whose trajectory is given by the hyperbola approaches the speed of light. But then we have already identified $\beta$ with $v/c$) Now, at $\beta = 1$, $x = ct$. And so $v = x/t = c.$ Or $v/c = 1.$ After that he writes $\beta = v/c$, perhaps because $v/c$ is a ratio and is dimensionless just like $\beta.$

 

[PeterDonis]

Let $\gamma = \cosh(\phi)$ and $\beta = \tanh(\phi),$

Note that this is a good example of "to get a solution to a problem, it helps to already know the answer". The two functions that appear in the previous equation are $\cosh (\phi)$ and $\sinh (\phi)$. So why is he defining symbols for $\cosh (\phi)$ and $\tanh (\phi)$? Obviously because he already knows that $\tanh (\phi)$ is going to turn out to be $v / c$.

Since

$$-c^2t^2+x^2=k$$
$$\gamma^2-\beta^2 \gamma^2 = 1.$$

Then if $\beta$ approaches 1, $-c^2t^2+x^2$ approaches 0. (I have no idea how he relates these two equations.

Since the transformation in question is equivalent to parameterizing the hyperbola by $\phi$, the obvious way to relate the equations is $x = \sqrt{k} \cosh (\phi) = \sqrt{k} \gamma$ and $ct = \sqrt{k} \sinh (\phi) = \sqrt{k} \beta \gamma$. Which, again, might seem more obvious if he had assigned a symbol directly to $\sinh (\phi)$.

However, as you have observed, this doesn't seem to help with understanding the limit $\beta \to 1$, since as $\beta \to 1$ along the hyperbola (corresponding to $\phi \to \infty$), $- ct^2 + x^2$ remains constant at $k$, it doesn't approach $0$. The limit he is implicitly trying to take here involves changing hyperbolas when you change $\beta$--higher values of $\beta$ correspond to hyperbolas with smaller values of $k$, which are therefore closer to the asymptotes that are the null lines $t = x$ and $t = - x$. But nothing in the equations he's given (or at least the ones you've shown us so far) has anything to do with taking that limit. So either your professor is doing some sleight of hand here, or he has given more information or equations that you haven't told us about.

 

[Orodruin]

However, as you have observed, this doesn't seem to help with understanding the limit $\beta \to 1$, since as $\beta \to 1$ along the hyperbola (corresponding to $\phi \to \infty$), $- ct^2 + x^2$ remains constant at $k$, it doesn't approach $0$. The limit he is implicitly trying to take here involves changing hyperbolas when you change $\beta$--higher values of $\beta$ correspond to hyperbolas with smaller values of $k$, which are therefore closer to the asymptotes that are the null lines $t = x$ and $t = - x$. But nothing in the equations he's given (or at least the ones you've shown us so far) has anything to do with taking that limit. So either your professor is doing some sleight of hand here, or he has given more information or equations that you haven't told us about.

... or is misunderstanding something himself. It would not be unheard of. The value of $\beta$ and the value of $k$ are independent. Regardless of $\beta$, the hyperbola for any $k$ is invariant. Even negative $k$ or $k = 0$ (the latter case not actually being a hyperbola but the light cone of the origin).

 

[Orodruin]

We want to preserve the dot product (actually the hyperbola):

$$-c^2t^2+x^2=k.$$

This suggests that the transformation that would preserve the hyperbola can be represented by

$$\cosh^(\phi) - \sinh^2(\phi) = 1.$$

Let $\gamma = \cosh(\phi)$ and $\beta = \tanh(\phi),$ then

$\gamma^2 - \beta^2 \gamma^2 = 1.$

'Now what is the nature of $\beta$?' asks the professor.

Well, it should be dimensionless since you cannot add a physical quantity ($\beta^2$) to a pure number (1):

$\gamma = \frac{1}{\sqrt{1-\beta^2}}.$

He goes on to write

Since

$$-c^2t^2+x^2=k$$
$$\gamma^2-\beta^2 \gamma^2 = 1.$$

Then if $\beta$ approaches 1, $-c^2t^2+x^2$ approaches 0. (I have no idea how he relates these two equations. It could be that if $\beta$ is the ratio $v/c$, then $\beta$ approaching 1 means that the particle whose trajectory is given by the hyperbola approaches the speed of light. But then we have already identified $\beta$ with $v/c$) Now, at $\beta = 1$, $x = ct$. And so $v = x/t = c.$ Or $v/c = 1.$ After that he writes $\beta = v/c$, perhaps because $v/c$ is a ratio and is dimensionless just like $\beta.$

Is this the actual verbatim argument presented?

 

[PeterDonis]

He goes on to write

Where is he writing this? Is it somewhere online that you can give a link to?

 

[Orodruin]

For the record, this is how the argument should go:

We wish to find a linear transformation
$$
ct' = A ct + B x, \quad x' = C ct + D x
$$
such that $(ct)^2 - x^2$ is invariant, i.e.,
$$
(ct')^2 - x'^2 = (A^2 - C^2) (ct)^2 - (D^2 - B^2) x^2 + 2(AB - CD) ctx = (ct)^2 - x^2
$$
for all $x$ and $t$. This is only possible if
$$
A^2 - C^2 = 1, \quad D^2 - B^2 = 1, \quad AB - CD = 0.
$$
The first two relations here can be parametrized as
$$
A = \cosh\theta, \ C = \sinh\theta, \ D = \cosh\phi, \ B = \sinh\phi
$$
for some $\theta$ and $\phi$. From the last relation then immediately follows
$$
\cosh\theta \sinh\phi - \sinh\theta\cosh\phi = \sinh(\theta - \phi) = 0
$$
which implies that $\theta = \phi$. We therefore have all of the transformation coefficients depending on a single variable $\theta$, which has to be determined. Assuming that the primed frame is the rest frame of an object at its origin that moves with speed $v$ in the unprimed frame, then $x = vt$ and $x' = 0$ for this object and therefore
$$
x' = x \cosh\theta + ct \sinh\theta = t(v \cosh\theta + c \sinh\theta) = 0.
$$
This is only possible for $t \neq 0$ if
$$
v = - c\tanh\theta.
$$
Using $\beta = v/c$ and defining $\gamma = \cos\theta$, this may be rewritten as $\beta = -\tanh\theta$ and therefore
$$
1 = \cosh^2\theta - \sinh^2\theta = \gamma^2 (1 - \beta^2) \quad \Longrightarrow \quad
\gamma = \frac 1{\sqrt{1-\beta^2}}.
$$
Summarizing, we therefore have
$$
ct' = \gamma(ct - \beta x), \quad x' = \gamma(x - \beta ct),
$$
which are the Lorentz transformation in standard setup.

 

[Samama Fahim]

I found the source of his lecture (he was writing this on the board):

https://books.google.com.pk/books?i...gy momentum vector special relativity&f=false

p. 17-19

 

[PAllen]

A point about pedagogy, IMO it is not clear why one should posit anything like the invariance of $(ct)^2-x^2$. In contrast, invariance of lightspeed is experimentally justified; or it can be derived that the principle of relativity, homogeneity and isotropy imply some invariant speed (which may be infinite), and then experiment again suggests which is the case. Did I miss an argument in this thread as to why one should assume 'hyperbolic invariance' as principle?

 

[Orodruin]

A point about pedagogy, IMO it is not clear why one should posit anything like the invariance of $(ct)^2-x^2$. In contrast, invariance of lightspeed is experimentally justified; or it can be derived that the principle of relativity, homogeneity and isotropy imply some invariant speed (which may be infinite), and then experiment again suggests which is the case. Did I miss an argument in this thread as to why one should assume 'hyperbolic invariance' as principle?

$(ct)^2-x^2$ being invariant implies the speed of light being invariant.

 

[Samama Fahim]

$(ct)^2-x^2$ being invariant implies the speed of light being invariant.

Please read the source I gave the link to in post # 26 or 28.

 

[Orodruin]

Please read the source I gave the link to in post # 26 or 28.

Huh? What do you mean by this? How is it connected to what I said which was a statement regarding the relation between the invariance of the spacetime interval amd the invariance of the speed of light?

 

[Samama Fahim]

Huh? What do you mean by this? How is it connected to what I said which was a statement regarding the relation between the invariance of the spacetime interval amd the invariance of the speed of light?

It's not.

 

[vanhees71]

That's true. In the usual way of deriving the LT, going back to Einstein's famous paper of 1905, is to use his "two postulates", which is (a) the special principle of relativity, i.e., the existence and indistinguishability of inertial reference frames and (b) the independence of the speed of light from the relative velocity between the light source and any inertial observer. In addition one tacitly assumes that for any inertial observer space is a 3D Euclidean affine space (implying its symmetries, i.e., translation and rotation symmetry) and the homogeneity of time (translation invariance in time).

First from the special principle of relativity we find that a free particle is moving with constant velocity with respect to any inertial frame of reference, and this implies that the transformation between two reference frames must be linear.

Now we consider the special case that we keep the directions of the spatial Cartesian bases the same for both frames of reference and consider only boosts in $x$ direction, i.e., then
$$c t'=A c t + B x, \quad x'=C ct + D x, \quad y'=y, \quad z'=z. \qquad (1)$$
We have assumed without loss of generality that the origins of space and time in both frames are chosen to be the same (if not, you can just redefine the coordinates by a time or space translation, which doesn't change anything because of the assumed homogeneity of time and space).

Now the 2nd postulate tells us that the wave front of a spherical em. wave switched on at $t=0$ from a source located at $\vec{x}=0$ obeys
$$c^2 t^2-\vec{x}^2=0, \qquad (2)$$
and the same must hold in $\Sigma'$, i.e., from (2) it necessarily follows also
$$c^2 t^{\prime 2}-\vec{x}^{\prime 2}=0,$$
i.e., there must be some factor $\alpha$ such that
$$c^2 t^2 - \vec{x}^2=\alpha (c^2t^{\prime 2}-\vec{x}^{\prime 2}).$$
Plugging in (1) you find
$$c^2 t^2 - \vec{x}^2=\alpha [(A c t + B x)^2 - (C ct + D x)^2-y^2-z^2]. \qquad (3)$$
Since this must hold for all $\vec{x}$ comparing the coefficients of $y^2$ and $z^2$ on both sides of the equation, it follows $\alpha=1$. So (3) reads with $\alpha=1$
$$c^2 t^2 -x^2 = (A^2-C^2) c^2 t^2 + 2 (AB-CD) ct x + (B^2-D^2) x^2.$$
Since this must be true for all $(ct,x)$ you find
$$A^2-C^2=1, \quad AB-CD=0, \quad B^2 - D^2=-1.$$
The rest then follows as shown by @Orodruin in #25.

 

[throw]

On p. 18, Robinson writes:
" corresponds to This corresponds to a vector with only a temporal component and no spatial component."

Why is that?

He explains why on the page, referencing equation (1.74). If you apply (1.75) to a four-vector and use $\beta = 0$ (what this implies for $\sinh$ and $\cosh$ is on p.18 explicitly too) you can see this explicitly.

 

[Samama Fahim]

He explains why on the page, referencing equation (1.74). If you apply (1.75) to a four-vector and use $\beta = 0$ (what this implies for $\sinh$ and $\cosh$ is on p.18 explicitly too) you can see this explicitly.

When $\beta = 0$, the transformation matrix we have, using 1.78, is the identity matrix. Applying this matrix to a four vector should leave this vector unchanged. Shouldn't it? And if this 4-vector has non-zero spatial components, they should be left untouched as well.