I Is this the only form of the Minkowski metric?

[kent davidge]

The Minkowski metric for inertial observers reads $ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2$. Is there a way to show that if it had off diagonal terms, the inertial observers would not see light traveling with the same speed?

 

[DrGreg]

The Minkowski metric for inertial observers reads $ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2$. Is there a way to show that if it had off diagonal terms, the inertial observers would not see light traveling with the same speed?

If you have an equation for the metric in a particular coordinate system, you can find the coordinate speed of light by putting $ds = 0$. For example, to find the coordinate velocity of light along the $x$-axis, put $ds = dy = dz = 0$ and solve to find two possible values for $dx/dt$.

You might like to try it with$$ds^2 = -(c^2-w^2) dt^2 + 2 w\,dt\,dx + dx^2 + dy^2 + dz^2$$

 

[kent davidge]

you can find the coordinate speed of light by putting $ds = 0$

and it seems intuitive that if the metric does not have that "inertial" form, the speed will not be $c$ as is the case in

You might like to try it with$$ds^2 = -(c^2-w^2) dt^2 + 2 w\,dt\,dx + dx^2 + dy^2 + dz^2$$

where I found $v = -w \pm c$ along x-axis, and $v = \pm \sqrt{c^2 - w^2}$ along the y and z axis.

But this does not show us that surely light will not travel at $c$ if the metric has off diagonal elements.

 

[PeterDonis]

Is there a way to show that if it had off diagonal terms, the inertial observers would not see light traveling with the same speed?

"Had off diagonal terms" is too vague. You need to make some kind of assumption about the coordinates, then show what the line element of Minkowski spacetime would look like in those coordinates (and show that there are off diagonal terms), and then compute what the coordinate speed of light would be. You might need to do this multiple times to cover all the possible ways that the line element could have off diagonal terms.

 

[DrGreg]

You may find it useful to know something about the diagonalisation of quadratic forms, and Sylvester's law of inertia.

 

[kent davidge]

You may find it useful to know something about the diagonalisation of quadratic forms, and Sylvester's law of inertia.

Thanks, I will look them up when I have the time. Can you tell what they imply? Maybe that light will always travel with speed different than $c$ if off diagonal elements are present?

 

[Orodruin]

The coordinate speed of light is rather uninteresting from a physics point of view. You can make it whatever just by rescaling your coordinates.

 

[DrGreg]

The Minkowski metric for inertial observers reads $ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2$. Is there a way to show that if it had off diagonal terms, the inertial observers would not see light traveling with the same speed?

Thanks, I will look them up when I have the time. Can you tell what they imply? Maybe that light will always travel with speed different than $c$ if off diagonal elements are present?

It seems to me that you might be making some assumptions about the metric and the coordinates, but you haven't explicitly stated what they are. For example:

• Are you assuming the spacetime is flat?
• Are you assuming that any particle at rest in the coordinate system is moving inertially (i.e. following a geodesic)?
• Are you assuming that all the components of the metric (in this coordinate system) are constant (i.e. are independent of coordinates, the same at every event)?

 

[kent davidge]

Im considering assumptions #1 and #2 and trying to see if they imply what assumption #3 states plus I am trying to see if they allow for off diagonal components in the metric.

 

[Nugatory]

Im considering assumptions #1 and #2 and trying to see if they imply what assumption #3 states plus I am trying to see if they allow for off diagonal components in the metric.

You can always find off-diagonal components, just by choosing appropriate coordinates. Consider ordinary boring two-dimensional Euclidean space, where we write the metric in Cartesian coordinates as $ds^2=dx^2+dy^2$ and in polar coordinates as $ds^2=dr^2+r^2d\theta^2$...

But we could try using coordinates $\alpha$ and $\beta$ defined by $\alpha=y$ and $\beta=y^2-x$. Write the metric in these coordinates (the easiest way to do this is to apply the tensor transformation law to the metric in Cartesian coordinates) and we will find off-diagonal elements. These off-diagonal elements have no physical significance, they’re just telling us that we’ve chosen a coordinate system in which the coordinate axes are not everywhere orthogonal.

 

[kent davidge]

You can always find off-diagonal components

These off-diagonal laments have no physical significance, they’re just telling us that we’ve chosen a coordinate system in which the coordinate axes are not everywhere orthogonal.

Does this mean that inertial observers can use coordinate systems where the metric has off diagonal elements?

 

[Nugatory]

Does this mean that inertial observers can use coordinate systems where the metric has off diagonal elements?

Any observer can use any coordinate system they want, any time and anywhere. So yes, an inertial observer can choose to use coordinates in which the metric has off-diagonal elements.

 

[kent davidge]

Any observer can use any coordinate system they want, any time and anywhere

I don't understand. If the "inertial frame" condition is the statement that a free particle is seen with constant speed, then if I use a coordinate system in which the particle is accelerating, then this coordinate system is not representing well my inertial frame.

 

[Nugatory]

If the "inertial frame" condition is the statement that a free particle is seen with constant speed,

The (Newtonian) inertial frame condition is that there exists a coordinate system in which a free particle moves at constant speed. Nothing requires that you use that coordinate system if you don’t want to.

 

[PeterDonis]

this coordinate system is not representing well my inertial frame

Yes, which means you probably won't want to use that coordinate system if you're interested in representing well your inertial frame. But that in no way means that coordinate system is not valid.

If what you are interested in is using coordinates that represent well your inertial frame, then I don't understand the point of this thread. If you're in flat Minkowski spacetime and moving inertially, and you want to represent well your inertial frame, obviously you're going to use standard Minkowski inertial coordinates. So why would you even ask about other coordinate systems?

 

[kent davidge]

If you're in flat Minkowski spacetime and moving inertially, and you want to represent well your inertial frame, obviously you're going to use standard Minkowski inertial coordinates. So why would you even ask about other coordinate systems?

Because perhaps there were other coordinate systems equaly good in representing the inertial frame. And I was asking whether they existed or not.

obviously you're going to use standard Minkowski inertial coordinates

Why is that obvious for you?

 

[Orodruin]

Because perhaps there were other coordinate systems equaly good in representing the inertial frame.

The Minkowski coordinates is the inertial frame so clearly it cannot be represented in any other way.

 

[PeterDonis]

perhaps there were other coordinate systems equaly good in representing the inertial frame. And I was asking whether they existed or not.

Then that's what you should have asked in the OP. Instead you asked, basically, what would happen if standard Minkowski coordinates had off diagonal terms. Which is like asking what would happen if two plus two equaled five. It's logically inconsistent.

 

[Nugatory]

There is some ambiguity here because "Minkowski metric" can refer to the metric tensor for Minkowski spacetime, or to the components of that tensor when written in Minkowski coordinates.

 

[PeterDonis]

There is some ambiguity here because "Minkowski metric" can refer to the metric tensor for Minkowski spacetime, or to the components of that tensor when written in Minkowski coordinates.

In the OP, it clearly refers to the latter since the line element is written down explicitly.

 

[kent davidge]

The Minkowski coordinates is the inertial frame so clearly it cannot be represented in any other way

But @Nugatory says in post #14 that they are just one possibility, that you can employ other coordinate systems and your frame is still inertial.

 

[PeterDonis]

@Nugatory says in post #14 that they are just one possibility, that you can employ other coordinate systems

Yes, but then you threw in the extra requirement that the coordinates have to "represent well" your inertial frame. In which case, as I said, this whole thread is pointless since there is only one coordinate chart that does that.

and you frame is still inertial.

What you really mean here is that your state of motion is inertial. Your state of motion is not the same as the coordinates (frame) you choose. You can use any coordinates you want, regardless of your state of motion. That was the point @Nugatory was making.

But if you insist, as you did (but only after the fact, not in the OP to this thread), that your coordinates have to "represent well" your "inertial frame", what you are saying is that your coordinates have to be well adapted to your state of motion. And if you are in flat Minkowski spacetime and your state of motion is inertial, there is only one coordinate chart that satisfies that requirement.

 

[PeterDonis]

perhaps there were other coordinate systems equaly good in representing the inertial frame. And I was asking whether they existed or not.

The response that only standard Minkowski coordinates meet this requirement does make some assumptions about what it means for a coordinate chart to "represent well" a particular observer's state of motion. Can you see what they are? Are they the same assumptions you are making about what "represent well" means?

 

[kent davidge]

Can you see what they are?

Yes, I guess what they must be.

Are they the same assumptions you are making about what "represent well" means?

No, as you said, I just tried to see if other coordinate systems were possible, without considering additional assumptions regarding the behaviour of nature in inertial frames.

 

[PeterDonis]

I just tried to see if other coordinate systems were possible, without considering additional assumptions

But "represent well" is an additional assumption. Are you making it or aren't you? And if so, what do you mean by it?

 

[kent davidge]

But "represent well" is an additional assumption. Are you making it or aren't you?

Ah, yes, I am

And if so, what do you mean by it?

By representing well, I have in mind that this coordinate system should give us constant velocity for a free particle.

 

[PeterDonis]

By representing well, I have in mind that this coordinate system should give us constant velocity for a free particle.

Ok. Have you considered what mathematical conditions this implies for the metric coefficients? Hint: consider what the geodesic equation looks like.

 

[kent davidge]

Have you considered what mathematical conditions this implies for the metric coefficients? Hint: consider what the geodesic equation looks like.

That was one issue I was facing when I started the thread. The equations will be $$\frac{d^2 x^\alpha}{d \tau^2} = 0$$ but this does not imply that the metric is constant or does not have off diagonal components. On the other hand, if we start out assuming that the metric components are constant, then we are lead to the above equation, and the particle is seen with constant velocity. This leaves open the possibility for off diagonal terms, though.

 

[kent davidge]

Hei, taking a look back at @DrGreg post #2, it seems that the additional assumption that should be made is that light travels with the same speed in all directions. This is enough to kill off the non diagonal terms in the metric.

 

[PeterDonis]

The equations will be
$$
\frac{d^2 x^\alpha}{d \tau^2} = 0
$$

but this does not imply that the metric is constant or does not have off diagonal components.

They do if you put back the crucial terms you left out:

$$
\frac{d^2 x^\alpha}{d \tau^2} + \Gamma^\alpha{}_{\beta \delta} \frac{d x^\beta}{d \tau} \frac{d x^\delta}{d \tau} = 0
$$

But, you say, those terms in the Christoffel symbols vanish in Minkowski coordinates! Yes, they do, but once more, if all you were concerned with were Minkowski coordinates, there would be no point in this thread! The whole point is that you want to investigate other possibilities for the coordinates, which means you have to include those other terms.

Once you include those other terms, and then impose your "represents well" requirement, which is that $dx^\alpha / d \tau$ must be constant for a geodesic, so that $d^2 x^\alpha / d \tau^2 = 0$, then you can see that we must have

$$
\Gamma^\alpha{}_{\beta \delta} \frac{d x^\beta}{d \tau} \frac{d x^\delta}{d \tau} = 0
$$

for all possible combinations of indices $\beta$ and $\gamma$. What does that imply about the metric? (Remember that the Christoffel symbols $\Gamma^\alpha{}_{\beta \delta}$ are constructed from derivatives of the metric.)

 

[PeterDonis]

the additional assumption that should be made is that light travels with the same speed in all directions. This is enough to kill off the non diagonal terms in the metric.

It's certainly sufficient, but it's stronger than necessary. Try considering my previous post first. (Hint: the geodesic equation applies to null geodesics, that light travels on, as well.)

 

[kent davidge]

for all possible combinations of indices $\beta$ and $\gamma$

but what if say, $dx^1 / d\tau = 0$? Then $\Gamma^\alpha{}_{1 \delta}$ could be different from zero.

 

[PeterDonis]

what if say, $dx^1 / d\tau = 0$? Then $\Gamma^\alpha{}_{1 \delta}$ could be different from zero

Free particles can travel in all directions, and you must meet the conditions for all possible 4-velocities for a free particle. That means you can't pick out any particular component of $dx / d\tau$ and say it's zero. You must allow for the possibility of all of them being nonzero (because there are free particle 4-velocities for which all of them are nonzero).

 

[kent davidge]

Free particles can travel in all directions, and you must meet the conditions for all possible 4-velocities for a free particle. That means you can't pick out any particular component of $dx / d\tau$ and say it's zero. You must allow for the possibility of all of them being nonzero (because there are free particle 4-velocities for which all of them are nonzero).

Ah, I got it. But since the $\Gamma$ are a sum of derivatives of the metric, wouldn't $\Gamma = 0$ only imply that the sum of these terms is zero? (And not necessarily the terms themselves, that is, the derivatives).

 

[PeterDonis]

since the $\Gamma$ are a sum of derivatives of the metric, wouldn't $\Gamma = 0$ only imply that the sum of these terms is zero?

For a single $\Gamma$, yes, that's all you could conclude. But we have that all of the $\Gamma$ are zero.