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A common way of constructing a pseudovector p is by taking the cross product of two vectors a and b:


p = a × b.


As x goes to −x, y to −y and z to −z, a and b go to −a and −b (by the definition of a vector), but p clearly does not change.



Is it right to say that every vector is a pseudovector as per the above stated defination,as we can represent any vector as a cross product of two vectors?

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No, every vector is not a pseudovector. A position vector is a classic example example of a non-pseudovector also known as polar vectors, because it will change it's sign upon inversion of the coordinate axes. A pseudovector won't, like your vector p there.


Usually, pseudovectors arise from descriptions of some sort of rotation, like the vorticity in a fluid, or angular momentum, or torque.

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What i am not able to understand is that even is case of a given position vector(say p), I can represent that specific position vector as cross product of two other vectors. Now under inversion,the cross product of these two vectors will remain same. Won't this make our given vector p a pseudovector?

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I think Bignose misunderstood your question. He is saying that "not every vector is a pseudo-vector". It is true that every pseudo-vector can be represented as the cross product of two vectors. It is, perhaps, more accurate to represent a pseudo-vector as an anti-symmetric 3 by 3 matrix.

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Halls, I didn't misunderstand anything. It is a driect response to


Is it right to say that every vector is a pseudovector


To which the answer is definately no.


I don't think that you can decompose any given vector into two crossed vectors. Certainly not uniquely, look at the cross product in Cartesion coordinates:


Let w = u x v, then the components of w are going to be given by:


[math]w_x = u_y v_z - u_z v_y[/math]

[math]w_y = u_z v_x - u_x v_x[/math]

[math]w_z = u_x v_y - u_y v_x[/math]


given that the components of w are known, you still have 6 unknowns and only 3 equations.


OK, you say, so, let's fix one of u or v, well, now the fixed one -- let's just pick v -- is a regular vector. That is, it will change sign under coordinate inversion. But, the other one is going to have to be a pseudovector so we get the correct w.


And, the now the properties of pseudovectors apply, namely:


[vector] x [pseudovector] = [vector], so we're back to where we started, with a regular vector.


So, again, I say (without confusion), no, not all vectors are pseudovectors.

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