# A Are 4-vectors ever measured without coordinates?

1. Aug 15, 2015

### Ken G

We know that 4-vectors are invariants, in the sense that they have the same meaning in all reference frames/coordinate systems. We know they transform by the Lorentz transformation in SR, and have an invariant Minkowski norm (let's not bring in GR at this point unless it becomes necessary). It seems to me that any time we wish to measure or determine experimentally a given 4-vector, we must do it by determining its components in some reference frame/coordinate system (I'm not trying to parse the differences between a reference frame and a coordinate system unless it becomes necessary to do so). So my question is, is it ever possible to measure or determine experimentally a 4-vector, without first choosing a reference frame/coordinate system?

2. Aug 15, 2015

### Orodruin

Staff Emeritus
I do not understand why you have this concern with 4-vectors specifically. The counter question would be "How do you measure or determine a 3-vector experimentally without choosing a frame of reference?" If you can answer that question, the answer is likely to be generalisable to 4-vectors.

3. Aug 15, 2015

### PAllen

Measurements are made with devices. Then we describe the set up of the devices, e.g. my calorimeter was at rest with respect to my particle source. The beam direction was between the emitter locked on my lab table and my target area. You have certainly picked out part of a frame basis in such a set up (e.g. the calorimeter's 'existence' picks a timelike basis vector). My lab as a whole picks out a moment to moment local spatial slice. So, I have a timelike basis and an orthogonal local spatial slice, i.e. a local foliation. I need not do more. I don't have to pick any specific spatial basis (just talk in terms of the direction from by source table to my target area). I certainly don't need to specify coordinates. Of course, it might make it easier describe in my lab notebook if I specified standard local coordinates for my lab that I used for a series of experiments. However, convenience is not necessity.

4. Aug 15, 2015

### vanhees71

To summarize PAllen's excellent posting #3: There's no way measuring anything without a frame of reference since your meaurement apparati define such a frame of reference.

5. Aug 15, 2015

### Ken G

Yet you do need to do that much-- you do need to choose a reference frame. So it sounds like your answer to my question is "no."
I didn't ask if you needed to pick a specific basis, only if you did have to at some point pick a basis. The reason this is important is, we say that 4-vectors don't depend on the basis we pick, but that is not the same thing as saying they have nothing do with picking a basis. My point here is that when we pick a basis, we are in effect choosing a language that we will use to quantify what is happening. So choosing a basis is like choosing a language, say French or English. We may be saying the same thing in French or English, and therefore what we are saying is "language independent", but this does not mean we can say it without picking a language.

6. Aug 15, 2015

### Ken G

The reason is because we say that 4-vectors do not depend on our reference frame/coordinates. I am making the point that "does not depend on" means something quite a bit different from "has nothing to do with."

7. Aug 15, 2015

### vanhees71

Sure, how else should you measure and describe the location of the pixels in your detector, which after all lets you do the proper analysis of which particle hit with which momentum this detector? Nowadays the data taking is fully automatic of course, and the computers just write out the corresponding data. To do so you need to define a frame of reference to store all these particle locations and times (or more precisely said the interaction events of the particle leting the detector "fire" and write out the information to the storage).

8. Aug 15, 2015

### Ken G

Right. Elsewhere in the forum I attempted to make this same point in regard to what are often called "physical explanations." I said that to give physical explanations, we choose a language in which to give that explanation, which in physics is tantamount to choosing a reference frame/coordinatization. It doesn't matter which language we pick, but we must pick a language. If instead, we express the answer entirely in terms of coordinate-free forms of some set of invariants, we cannot test what we are claiming, because as we just saw-- we can't do the observations. When I made that exact point, it was claimed I was "stubbornly clinging" to a false idea. I'm glad this thread has confirmed what I was saying, which I believe was simply being misunderstood.

9. Aug 15, 2015

### vanhees71

Well, on the other hand, at the end, everything really physically meaningful is independent of the choice of your reference frame, i.e., it can be expressed in terms of invariants. That's why, e.g., one measures invariant cross sections in high-energy experiments etc.

10. Aug 15, 2015

### Ken G

Yes, and anything you can say is pretty much independent of whether you say it in English or French, correct? Yet you need a language to say anything, and what you say will sound very different in different languages.

I realize that physics must be expressed in terms of invariants, this is all part of being an objective science. My point is that a science based on objectivity still does not escape the need to have a subjective language in which to express those things that end up being invariant of the language. And so it is with physical explanations-- they are in a language, often involving convenient choices of reference frame/coordinates. The explanation does depend on the language, in the sense that the same explanation will sound quite different in different languages. Underneath the explanation are invariants, and we can use the language to access the invariants, but there always requires a choice of language, and then there can be an explanation. Relativity is the place where we come to terms with the fact that the explanations do not sound unique, even though underneath the explanations, we have the same invariants. This comes up when we try to give physical reasons why things happen, like the reason the rope breaks in the Bell spaceship problem. I regard this as a very uncontroversial claim, so I'm checking in to make sure I'm not missing something, which is why I asked the above question.

11. Aug 15, 2015

### Orodruin

Staff Emeritus
We say this about 3-vectors too - they do not depend on whether you pick your x-axis to be north or west. It is nothing special for 4-vectors.

[[Moderator note: Topic closed, it has already been discussed elsewhere.]]

Last edited: Aug 19, 2015