# Allowable Coordinate Transformations?

1. Jul 30, 2011

### mmmboh

I've studied classical physics and never heard this before until recently...the allowable coordinate transformations for classical mechanics are rotations and translations. Could someone explain why this is so? What makes these "allowable" (I know they are orthogonal transformations).

2. Jul 30, 2011

### RedX

That's just a consequence of using vectors. If you have a vector equation, such as Newton's law, then rotations and translations result in the same equation.

If in addition you make other assumptions, such as homogeneous forces, you can stretch and dilate. For example -kx=mx'', you can make transformations y(x)=cx and still get -ky=my'' for a constant c.

3. Jul 30, 2011

### mmmboh

So allowable coordinate transformations just means coordinate transformations that leave the equations the same?

What about Galilean boosts? F=ma either way.

I mean if Lorentz boosts are allowable in SR, why aren't Galilean boosts allowable boosts allowable in CM?

Last edited: Jul 30, 2011
4. Jul 30, 2011

### RedX

You have to make some assumptions with Gallilean boosts. For example, if you have a force that depends on velocity, say a drag force instead of a spring force:

-kx'=mx''

The transformation y=x+vt (where v is velocity, t is time) results in the equation:

-k(y'-v)=my''

which is not -ky'=my''

Now of course Gallilean boosts are allowable. What happens is that the force is not really -kx', but -k(x'-u), where u here is the velocity of the fluid. When you boost, both x' and u change by the same amount, and you get: -k(y'-u')=my'' which is the same as -k(x'-u)=mx''. But you have to make a statement about forces for this to be true.

5. Jul 30, 2011

### mmmboh

So why wouldn't we have this problem when using Lorentz boosts?

6. Jul 30, 2011

### RedX

I think that's an interesting historical question. The equations for electricity and magnetism, for example, are not the same when you make a Gallileo boost of the form y=x+vt. People tried to save it by introducing an ether fluid, and so when you boost you also have to boost the ether fluid. But ultimately people abandoned the notion that the laws of physics must remain the same under Gallilean boosts, and used Lorentz boosts instead.

So the requirement that the equations must remain the same under a Lorentz transformation greatly restricts what type of forces you can have. Just as you cannot have the equation:

-kx'=mx''

because it violates Gallileo boosts, and instead must modify it by:

-k(x'-u)=mx''

where u is an ether, requiring that the equations must be the same under Lorentz transformation eliminates many possible forces that people can conjure up.

In fact, it's even more than that. Requiring that the equations remain the same under Lorentz transformation actually restricts what type of particles you can have. There are only a couple of types: scalar particles, vector particles, tensor particles, spinor particles, and mixed tensor/spinor particles. So if you want to invent a new type of mathematical object to describe a particle instead of a vector or spinor, then it's actually very hard to do because Lorentz transformation takes away a lot of your freedom: you would have to find a new representation of the Lorentz group besides spinors and vectors, in the jargon of particle physics.

7. Jul 30, 2011

### Rasalhague

I think here "allowable" is short-hand for the property these transformations have of transforming one inertial coordinate system into another inertial coordinate system, i.e. another coordinate system of that special sort where Newton's laws of motion take their famous, simplest form. Of course, other transformations can be used, but not without modifying F = ma with other coordinate-dependent, "fictitious" forces.

8. Jul 30, 2011

### mmmboh

But Newton's laws are invariant under Galilean boosts, but kinetic energy is not...does it still make sense to call kinetic energy a scalar if this is an allowable coordinate transformation?

Thanks for the replies, good info.

9. Jul 30, 2011

### RedX

Kinetic energy would not be a scalar because of Galilean boosts. If you're standing still, a tree has very little kinetic energy. If you're in a moving car, the tree is moving very fast. However, the kinetic energy is a scalar in the sense that is remains invariant under rotations. So it doesn't matter which direction you're facing: the object will have the same kinetic energy. So if someone says kinetic energy is a scalar, they mean in the sense that it doesn't change under translations or rotations. But it will change under boosts - a tree has a lot of kinetic energy if the observer is moving really fast in a car.

10. Jul 31, 2011

### BruceW

Yeah, I think the idea is that when you boost, the energy of objects will become different, but energy is still conserved in the boosted frame.

11. Jul 31, 2011

### Staff: Mentor

Yes, energy is frame variant, but conserved.

In tensor terms energy is definitely NOT a scalar; it is a component of a vector.

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