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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 travelling 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##.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 travelling with the same speed?
and it seems intuitive that if the metric does not have that "inertial" form, the speed will not be ##c## as is the case inyou can find the coordinate speed of light by putting ##ds = 0##
where I found ## v = -w \pm c## along x-axis, and ##v = \pm \sqrt{c^2 - w^2}## along the y and z axis.You might like to try it with$$ds^2 = -(c^2-w^2) dt^2 + 2 w\,dt\,dx + dx^2 + dy^2 + dz^2$$
"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.Is there a way to show that if it had off diagonal terms, the inertial observers would not see light travelling 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?You may find it useful to know something about the diagonalisation of quadratic forms, and Sylvester's law of inertia.
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 travelling with the same speed?
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: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?
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##....Im considering assumptions #1 and #2 and trying to see if they imply what assumption #3 states plus Im trying to see if they allow for off diagonal components in the metric.
You can always find off-diagonal components
Does this mean that inertial observers can use coordinate systems where the metric has off diagonal elements?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.
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.Does this mean that inertial observers can use coordinate systems where the metric has off diagonal elements?
I dont 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.Any observer can use any coordinate system they want, any time and anywhere
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.If the "inertial frame" condition is the statement that a free particle is seen with constant speed,
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.this coordinate system is not representing well my inertial frame
Because perhaps there were other coordinate systems equaly good in representing the inertial frame. And I was asking whether they existed or not.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?
Why is that obvious for you?obviously you're going to use standard Minkowski inertial coordinates
The Minkowski coordinates is the inertial frame so clearly it cannot be represented in any other way.Because perhaps there were other coordinate systems equaly good in representing the inertial frame.
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.perhaps there were other coordinate systems equaly good in representing the inertial frame. And I was asking whether they existed or not.
In the OP, it clearly refers to the latter since the line element is written down explicitly.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.
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.The Minkowski coordinates is the inertial frame so clearly it cannot be represented in any other way
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.@Nugatory says in post #14 that they are just one possibility, that you can employ other coordinate systems
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.and you frame is still inertial.
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?perhaps there were other coordinate systems equaly good in representing the inertial frame. And I was asking whether they existed or not.
Yes, I guess what they must be.Can you see what they are?
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.Are they the same assumptions you are making about what "represent well" means?
But "represent well" is an additional assumption. Are you making it or aren't you? And if so, what do you mean by it?I just tried to see if other coordinate systems were possible, without considering additional assumptions