# [Special Relativity] Scalar Invariant under a Lorentz-transformation

• Athenian
In summary, @eranreches's solution for this question is based off of understanding how the quantities ##u^\alpha v^\beta## transform under the Lorentz transformations, and then concluding that the quantity ##u^\alpha v^\beta## is not a scalar invariant.
Athenian
Homework Statement
Given the components of two vector fields, ##u^\alpha, v^\beta##, show that ##u^\alpha v^\alpha = u^0 v^0 + u^1 v^1 + u^2 v^2 + u^3 v^3## is not a scalar invariant under a Lorentz-transformation.
Relevant Equations
Refer below ##\rightarrow##
"My" Attempted Solution

To begin, please note that a lot - if not all - of the "solution" is largely based off of @eranreches's solution from the following website: https://physics.stackexchange.com/questions/369352/scalar-invariance-under-lorentz-transformation.

With that said, below is my attempt to apply what @eranreches's did in his solution for this problem while trying to understand how the solution works. I would sincerely appreciate it if anyone in the community could help either solidify my understanding or perhaps correct any mistakes I may have made below. Thank you very much for all your time and assistance!

Since - as @eranreches's stated - the quantity ##u^\alpha v^\beta## has a tensor of rank ##(2,0)##, it can then be expressed by a ##4\times 4## matrix as seen below:

##\begin{pmatrix}u^0v^0&u^0v^1&u^0v^2&u^0v^3\\u^1v^0&u^1v^1&u^1v^2&u^1v^3\\u^2v^0&u^2v^1&u^2v^2&u^2v^3\\u^3v^0&u^3v^1&u^3v^2&u^3v^3\end{pmatrix}##

where the diagonal elements refer specifically to ##u^\alpha v^\alpha## (i.e. ##u^0 v^0, u^1 v^1, u^2 v^2, u^3 v^3##).

In the case of this question, the quantity ##u^\alpha v^\beta## would transform like below (also based off of @eranreches's answer):

$$u^{\alpha^\prime} v^{\beta^\prime}=\frac{\partial x^{\alpha^\prime}}{\partial x^\alpha}\frac{\partial x^{\beta^\prime}}{\partial x^\beta}u^\alpha v^\beta$$

but, specifically, for the diagonal elements this time around, we would get ##\Rightarrow##

$$u^{\alpha^\prime} v^{\alpha^\prime}=\frac{\partial x^{\alpha^\prime}}{\partial x^\alpha}\frac{\partial x^{\alpha^\prime}}{\partial x^\beta}u^\alpha v^\beta$$

After writing the above equations, the author came to the conclusion that "In general, [the above equation in the question] is not invariant under the Lorentz transformations".

However, what I have a hard time understanding is how does the above equation prove that "##u^\alpha v^\alpha = u^0 v^0 + u^1 v^1 + u^2 v^2 + u^3 v^3## is not a scalar invariant under a Lorentz-transformation"?
Is it because the equation for the diagonal elements right above didn't have ##u^\beta v^\beta## as equaling to ##u^\alpha v^\alpha## - as in, kind of similar to what is written below?

$$u^{\alpha^\prime}v_{\alpha^\prime}=\frac{\partial x^{\alpha^\prime}}{\partial x^\beta}u^\beta\frac{\partial x^{\gamma}}{\partial x^{\alpha^\prime}}v_\gamma=\frac{\partial x^{\gamma}}{\partial x^\beta}u^\beta v_\gamma=\delta^{\gamma}_{\beta}u^\beta v_\gamma=u^\beta v_\beta$$

To wrap up, if anyone could correct any of my potential mistakes or help clear up any confusion I have, I would sincerely appreciate it. Thanks in advance!

Perhaps a simpler solution is to produce a counterexample.

Note that the two quantities need not be different in all cases. But, there are cases where they are different.

Can you think of examples of both?

In general, it's not a good idea to dive into complicated algebra unless you need complicated algebra.

PeroK said:
Perhaps a simpler solution is to produce a counterexample.

Note that the two quantities need not be different in all cases. But, there are cases where they are different.

Can you think of examples of both?

In general, it's not a good idea to dive into complicated algebra unless you need complicated algebra.

Admittingly, after giving your question some thought, I am having an incredibly difficult time coming up with examples or cases of the two quantities being different or the same. However, when you asked that am I able to come up with examples of "both", do you refer to examples of the two quantities being invariant or not invariant?

Regardless of which one, I do indeed have a difficult time coming up with an example for either one of them. If possible, could you please provide a guiding example to help me better understand?
Thank you!

Athenian said:
Admittingly, after giving your question some thought, I am having an incredibly difficult time coming up with examples or cases of the two quantities being different or the same. However, when you asked that am I able to come up with examples of "both", do you refer to examples of the two quantities being invariant or not invariant?

Regardless of which one, I do indeed have a difficult time coming up with an example for either one of them. If possible, could you please provide a guiding example to help me better understand?
Thank you!
An invariant quantity is one that is the same in all reference frames. However, even if a quantity is not generally invariant, it might have the same value in two specific reference frames.

I think it's better if you think about the SR case yourself.

Let me suggest that you simplify the problem by considering only the ##t## and ##x## coordinates.

@Athenian my advice would be to leave this advanced vector and tensor notation for the time being. The post on stack exchange doesn't answer the question. He only points out the general notation of the covariant and contravariant components. This tensor algebra becomes a high mathematical hurdle to get over, and needs quite a bit of study in it's own right.

For SR, I.e. flat spacetime, you can avoid these mathematical difficulties by simply considering a four-vector to have four components. With the inner product defined with the negative for the ##t## or ##0## components.

Athenian
PeroK said:
@Athenian my advice would be to leave this advanced vector and tensor notation for the time being. The post on stack exchange doesn't answer the question. He only points out the general notation of the covariant and contravariant components. This tensor algebra becomes a high mathematical hurdle to get over, and needs quite a bit of study in it's own right.

For SR, I.e. flat spacetime, you can avoid these mathematical difficulties by simply considering a four-vector to have four components. With the inner product defined with the negative for the ##t## or ##0## components.

Thank very much for the reply, @PeroK! I sincerely appreciate the response.

After a day's worth of research online, below is what I came up with and managed to understand.

However, before I begin, allow me to do a quick concept check. Would an "inner product" be considered as two vectors multiplied by each other to produce a scalar "real number" quantity? If so, by multiply, would that be a dot product we're talking here or rather a cross-product instead? Maybe it's just me, but I felt like the websites I visited were especially vague on this matter.

After mulling over your suggestion of considering a "four-vector ... with the inner product define with the negative for the ##t## or ##0## components", I am still utterly confused and have no idea on how to logically go about doing that.

But, at the very least, below is what I have managed to gather for the past day:

To make ##u^\alpha v^\alpha## invariant - that is, to create a counter-example to the given problem - the following equation below should be able to show ##u^\alpha v^\alpha## 's invariance.
$$\begin{pmatrix} u^0 & u^1 & u^2 & u^3\end{pmatrix} \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & +1 & 0 & 0 \\ 0 & 0 & +1 & 0 \\ 0 & 0 & 0 & +1 \end{pmatrix} \begin{pmatrix} v^0 \\ v^1 \\ v^2 \\ v^3\end{pmatrix}$$

And, this same idea will apply to another reference frame (say, ##u^{\alpha '} v^{\alpha '}##) - hence proving that the counter-example equation ##u^\alpha v^\alpha = -u^0 v^0 + u^1 v^1 + u^2 v^2 + u^3 v^3## is indeed invariant:
$$\begin{pmatrix} u^{0'} & u^{1'} & u^{2'} & u^{3'}\end{pmatrix} \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & +1 & 0 & 0 \\ 0 & 0 & +1 & 0 \\ 0 & 0 & 0 & +1 \end{pmatrix} \begin{pmatrix} v^{0'} \\ v^{1'} \\ v^{2'} \\ v^{3'}\end{pmatrix}$$

However, note that since the shown equation in the original question does not have the negative sign before ##u^0 v^0##, one can thus not use ##\eta## - that is the below matrix

$$\begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & +1 & 0 & 0 \\ 0 & 0 & +1 & 0 \\ 0 & 0 & 0 & +1 \end{pmatrix}$$

for this: ##u^\alpha v^\alpha = u^0 v^0 + u^1 v^1 + u^2 v^2 + u^3 v^3##
Rather than ##\eta##, our equation right above will simply have the matrix identity operator ##\delta## used instead.

But, despite "knowing" all this, I am nevertheless confronted with the same question - how can I prove that the question's equation is not a scalar invariant under a Lorentz-transformation? Or, would my counter-example to disapprove the question's equation being invariant suffice?

I sincerely apologize for any inconvenience I may have caused due to the number of questions I have here. But, as you have already nicely done, I would greatly appreciate any help to clear up some of my confusion. Thank you!

-----------------------------------------------------
Update:
By what you said below:

PeroK said:
For SR, I.e. flat spacetime, you can avoid these mathematical difficulties by simply considering a four-vector to have four components. With the inner product defined with the negative for the ##t## or ##0## components.

Would the short answer given by the user named @WetSavannaAnimal in the provided URL perhaps sort of answer your given suggestion to me?
https://physics.stackexchange.com/q...s-not-invariant-under-galilean-transformation

You are seriously over-complicating this.

If we take the following coordinate transformation:

##t' = t, \ \ x' = -x##

In this case, we have both

##t^2 - x^2 = t'^2 - x'^2##

and

##t^2 + x^2 = t'^2 + x'^2##

So, it's not always the case that this quantity is different after a coordinate transformation.

If we extend this to 2D or 3D, then as long as we have ##t' = t##, any rotation of the spatial axes will again lead to equality.

To lose the invariance of ##t^2 + x^2 + y^2 + z^2## we must, therefore, do something with the time coordinate. The simplest case is the 1D Lorentz transformation:

##t' = \gamma(t - vx/c^2), \ \ x' = \gamma(x - vt)##

Let's assume that we have shown that the quantity ##t^2 - x^2## is invariant. I.e., we know that:

##t^2 - x^2 = t'^2 - x'^2##

If we also have

##t^2 + x^2 = t'^2 + x'^2##

Then, adding or subtracting these two equations gives:

##t^2 = t'^2## and ##x^2 = x'^2##

And that is not generally true for any choice of ##t, x##. I.e. for any position four-vector. Unless, of course, ##v= 0## and ##\gamma = 1##. E.g. take ##x = 0## and ##t \ne 0##.

In summary, what we have shown is:

1) If we do not change the time coordinate, then both the "+" and "-" quantities are invariant under that specific (spatial) coordinate transformation.

2) If we do a simple Lorentz boost in the x-direction, then the "+" quantity is not in general invariant for all postion four-vectors.

Note that to show a quantity is not invariant you only need to find a single counterexample. You don't have to prove that the quantity is never the same. In fact, there will often be special cases where it is the same.

That's why the post in stack exchange isn't really a proof of anything. What you really need is a single counterexample.

Athenian
Thank you for the clear and detailed explanation, @PeroK! I sincerely appreciate it.

After a lot more studying - hence the delayed reply - below is the final solution I came up with. I would sincerely appreciate it if you could confirm whether the answer is correct or not.

To begin with, I will be considering the below equation (in the perspective of the question) to be a Lorentz invariant (which I will be putting to the test later on below):
$$\vec{u^\alpha} \cdot \vec{v^\alpha} = \vec{u^\alpha}' \cdot \vec{v^\alpha}'$$

Please note that here, ##\vec{u^\alpha}' = L\vec{u^\alpha}## and ##\vec{v^\alpha}' = L\vec{v^\alpha}##, where ##L## stands for the Lorentz transformation of the vector(s).
Below is the following Lorentz transformation matrix (##L##) used in the last sentence:

$$L = \begin{bmatrix} \gamma (u^\alpha) & -u^\alpha \gamma (u^\alpha) & 0 & 0 \\ -u^\alpha \gamma (u^\alpha) & \gamma (u^\alpha) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$
and
$$L = \begin{bmatrix} \gamma (v^\alpha) & -v^\alpha \gamma (v^\alpha) & 0 & 0 \\ -v^\alpha \gamma (v^\alpha) & \gamma (v^\alpha) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$
where ##\gamma (u^\alpha) = \frac{1}{\sqrt{1- \frac{{(u^\alpha)}^2}{c^2}}}## and ##\gamma (v^\alpha) = \frac{1}{\sqrt{1- \frac{{(v^\alpha)}^2}{c^2}}}## respectively.

Continuing on, we can get ##\vec{(u^\alpha)}^T \vec{v^\alpha} =\vec{(u^\alpha)}^T L^T L \vec{v^\alpha} = \vec{(u^\alpha)}^T \eta \vec{v^\alpha}##.

Therefore, because of the above result, I concluded that ##\vec{u^\alpha} \cdot \vec{v^\alpha} \neq \vec{u^\alpha}' \cdot \vec{v^\alpha}'## - thus, leading me to believe that this example is a counterexample of the given vector equation provided in the question to be not invariant.

I really hope I'm not over-complicating my answer again this time around. Needless to say, I will appreciate any corrections or solution confirmation to my answer above.
Thank you for all your patience and effort in assisting me here! You have no idea how much it means to me to have someone knowledgeable in the subject to help me better understand the material. And for that, I sincerely thank you!

My counterexample is ##\gamma = 2, \ t =1, \ x = 1##.

That's the simplest form of a counterexample.

Athenian

## 1. What is a scalar invariant in the context of special relativity?

A scalar invariant is a physical quantity that remains unchanged under a Lorentz transformation, which is a mathematical transformation that relates measurements made by observers in different inertial reference frames. In other words, a scalar invariant does not depend on the relative motion of observers and is the same for all inertial frames of reference.

## 2. How is a scalar invariant different from a scalar quantity?

A scalar quantity is a physical quantity that has magnitude but does not have a direction. Examples of scalar quantities include mass, temperature, and time. On the other hand, a scalar invariant is a specific type of scalar quantity that remains the same under a Lorentz transformation. While all scalar invariants are scalar quantities, not all scalar quantities are scalar invariants.

## 3. What are some examples of scalar invariants in special relativity?

Some examples of scalar invariants in special relativity include space-time intervals (the difference in space and time coordinates between two events), proper time (the time measured by a clock in its own frame of reference), and rest mass (the mass of an object measured in its own frame of reference).

## 4. Why is the concept of scalar invariants important in special relativity?

The concept of scalar invariants is important in special relativity because it allows for the invariance of physical laws and principles. This means that the fundamental laws of physics, such as conservation of energy and momentum, hold true in all inertial frames of reference. Scalar invariants also allow for the prediction and understanding of relativistic effects, such as time dilation and length contraction.

## 5. How are scalar invariants used in practical applications of special relativity?

Scalar invariants are used in various practical applications of special relativity, such as in GPS technology. The time dilation effect, which is a result of the invariance of proper time, must be taken into account for accurate GPS calculations. Scalar invariants also play a role in particle accelerators, where the rest mass of particles is used to calculate their kinetic energy and the effects of relativistic speeds on their behavior.

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